Download Unit 8: Reactions

Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Unit 8:
Reactions
1.
Student Name: _____________Key_______________
Class Period: _3, 5, & 10_
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Unit 8 Vocabulary:
1. Coefficient: A number placed in front of a formula to balance a
chemical reaction.
2. Decomposition: A redox reaction in which a compound breaks up to
form two elements.
3. Double Replacement: A solution reaction in which the positive ion of
one compound combines with the negative ion of the other
compound to form a precipitate, and the other ions remain dissolved
in solution.
4. Law of Conservation of Charge: Charge may not be created or
destroyed by physical or chemical change. This is the basis for
writing chemical formulas and half-reactions, and for balancing
redox reactions.
5. Law of Conservation of Energy: Energy may not be created or
destroyed by physical or chemical change. This is the basis for
calculating the heat of reaction.
6. Law of Conservation of Mass: Matter may not be created or
destroyed by physical or chemical change. This is the basis for
balancing chemical reactions.
7. Mole Ratio: The whole-number ratio between components of a
balanced chemical reaction.
8. Oxidation: The loss of electron(s), causing the oxidation number of a
species to become more positive.
9. Precipitate: An insoluble solid that is formed in either a doublereplacement reaction or as an excess solute added to a saturated
solution.
10. Product: The substances that are formed by a chemical reaction,
shown on the right side of a chemical equation.
11. Reactant: The substances that are the beginning of a chemical
reaction, shown on the left side of a chemical equation.
12. Redox reaction: A reaction in which one element is oxidized and
another element is reduced.
13. Reduction: The gain of electron(s), causing the oxidation number of
a species to become more negative.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
14. Single Replacement: A redox reaction in which an element replaces
an ion within a compound.
15. Spectator Ion: An ion that does not participate in the overall
chemical reaction. In a redox reaction, the spectator ion is the ion
whose charge does not change. In a double replacement reaction,
they are the ions that remain dissolved in solution.
16. Stoichiometry: The mathematics of mole relationships.
17. Synthesis: A redox reaction in which two elements combine to form
a compound.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Unit 8 Homework Assignments:
Assignment:
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Date:
Due:
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Notes page:
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Lecture Key
Unit 8: Reactions-Key
Topic:
Regents Chemistry ’14-‘15
Mr. Murdoch
Chemical Equations
Objective: What is a chemical equation?
Chemical Equation: A chemical equation is a symbolic representation of
a chemical reaction. The chemical equation includes:
1. the substances starting the process (reactants);
2. the substances being formed during the process (products);
3. the phases of matter for the substances during the process;
4. the number of moles of each substance in the process;
5. and the resulting change in energy of the entire process.
Reactants  Products
Coefficients are whole numbers placed in front of the substance symbols
to denote a mole ratio that functions within the Law of Conservation of
Mass.
HCl(aq) + Zn(s)  H2(g) + ZnCl2(aq)
(reactants)
(products)
This equation states that the reactant aqueous hydrochloric acid (HCl)
when combined (+) with solid zinc metal (Zn) undergoes a reaction ()
to produce hydrogen gas (H2) and aqueous zinc chloride (ZnCl2).
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Law of Conservation of Mass
Objective: How is the Conservation of Mass applied to equations?
Conservation of Mass:
 The Law of Conservation of Mass states that mass may not be created
or destroyed by physical or chemical changes.
 Any elements found on the reactants (left) side MUST be found on
the products (right) side, and the numbers of moles of those elements
MUST be equal on both sides.
Look at the first equation we used in this unit:
HCl(aq) + Zn(s)  H2(g) + ZnCl2(aq)
(reactants)
(products)
This equation breaks the Law of Conservation of Mass. Note that there
are unequal numbers of moles of H and Cl on both sides of the equation.
H + Cl(aq) + Zn(s)  H
(g)
+ Zn Cl
(aq)
We can’t just add matter, so we need to place coefficients to act as
multipliers in front of a substance’s formula:
HCl(aq) + Zn(s)  H2(g) + ZnCl2(aq)
We didn’t have to change anything with the zinc (Zn), as there is only
one mole (the 1 is not shown) of Zn on either side.
This states that 2 moles of HCl are required to react with 1 mole of Zn.
Watch YouTube Conservation of Mass video (1:25)
https://www.youtube.com/watch?v=mcnga-bbNXk
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Rules for Writing Equations
Objective: What are the rules for writing chemical equations?
Writing Chemical Equations:
If you are given the names in your equation, the steps for writing a
chemical equation are as follows:
1. Balance the Equation:
i. Write in PENCIL (you’ll make mistakes; we all do!);
ii. Write the coefficients in one element at a time;
iii. You may only add coefficients; you may NOT change any
chemical formulas when balancing the equation;
iv. Revise (remember, to use PENCIL) the coefficients where
necessary.
*Note: A coefficient is the number in front; a subscript is behind!
 When you write 2 Cl, that states there are TWO atoms of chlorine.
 When you write Cl2, that states there is ONE molecule of diatomic
(2 atoms) chlorine. Diatomic molecules of (Br2, I2, N2, Cl2, H2, O2,
& F2) exist whenever these elements are not in a compound with
another element.
In NaCl, there is one Cl-1 ion (due to Na’s+1 charge), but when Cl-1 is
separated from another compound, the two single Cl-1 ions covalently
share their combined 14 valence electrons as one diatomic Cl2 molecule.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
NaCl  Na + Cl
Again, the separated Cl-1 ions will form a diatomic molecule, which
“adds” matter:
NaCl  Na + Cl2
However, this throws off the coefficient balance.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Balancing Equations
Topic:
Objective: How do we balance the coefficients used in an equation?
Balancing an Equation:
Look at the equation of NaCl  Na + Cl2:
i. There is one Na on the left and one Na on the right. Na is
balanced…for the moment.
ii. There is one Cl atom on the left and two Cl atoms on the right.
Multiply them together (1 x 2 = 2):
 This means when balanced, there should be two chlorine
atoms on BOTH sides of the arrow.
iii. As there exists 2 Cl atoms on the right (products) side, place a
coefficient of in front of the entire NaCl formula on the left side:
 That makes
NaCl  Na + Cl2
iv. However, now look at Na on the right (products) side of the arrow.
When we added the 2 coefficient in front of NaCl, we not only
multiplied the Cl by 2, but we multiplied the Na by 2 as well.
v. So, add a coefficient of in front of the Na on the right (products)
side:
 That gives us 2 NaCl 
Na + Cl2
vi. Doing the math, there are 2 Na atoms & 2 Cl atoms on the left
(reactants) side, and 2 Na atoms & 2 Cl atoms on the right
(products) side. The equation (and reaction) is BALANCED!!!
Watch YouTube Bozeman Science Balancing Equations video (9:38)
https://www.youtube.com/watch?v=wxvERNlUdBQ
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Balancing an Equation example #1:
H2 + O2  H2O
i. There are two H’s on the left and two H’s on the right. OK…for
now.
ii. There are two O’s on the left, and only one O on the right. Multiply
(2 x 1 = 2) and place the resulting coefficient of in front of the
H2O on the right:
H2 + O2 
H2O
 There will now be two O’s on either side of the arrow, but
now there are four total H’s on the right.
iii. Solve this by placing a coefficient of in front of the H2 on the left:
H2 + O2  2 H2O
 There are now four H’s on BOTH sides, and two O’s on
BOTH sides.
iv. The equation is balanced!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Balancing an Equation example #2:
Al + O2  Al2O3
i. There is one Al atom on the left and two Al atoms on the right.
Multiply (1 x 2 = 2) and place the resulting coefficient of in front
of the Al on the left:
Al + O2  Al2O3
ii. There will now be two O’s on the left and three O’s on the right.
Multiply (2 x 3 = 6) which means there should be six O’s on both
sides.
iii. On the left side of the arrow there are two O’s. So, (2 x Y = 6), with
Y = 3. Therefore place a in front of the O2 on the left.
2 Al +
O2  2 Al2O3
iv. On the right side of the arrow there are three O’s (in Al2O3). Divide
6 by 3 and you get 2, so place a on the right side:
2 Al + 3 O2 
v.
Al2O3
 Uh-oh. Al is now wrong, as there are only two Al atoms on
the left, and a total of four Al atoms on the left.
(you used PENCIL, right?) the coefficient of on the
left side with a coefficient of :
Al + 3 O2  2 Al2O3
 There are now four Al’s and six O’s on either side of the
arrow.
vi. Balanced!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Balancing an Equation example #3:
SO3  SO2 + O2
i. Note here that O is found in THREE places; once on the left, twice
on the right.
ii. There is one S on both sides. S is balanced…for a bit.
iii. There are three O’s on the left side of the arrow, but a total of four
O’s on the right side of the arrow. Multiply (3 x 4 = 12) which
means there should be 12 atoms of O on either side when balanced.
Divide the three O’s on the left by 12, which leaves 4, so place a
coefficient of in front of the SO3:
SO3  SO2 + O2
 We now have 12 total O’s and 4 total S’s on the left.
iv. That means we have to add a coefficient of in front of the right
side SO2 to have 4 S’s on the right.
4 SO3 
v.
SO2 + O2
 This gives 8 total O’s in the SO2, but we need 12 total O’s on
the right.
So, (12 – 8 = 4), so place a coefficient of in front of the O2 (2 x
O2 = 4 O) to give a total of 12 O’s on the right.
4 SO3  4 SO2 +
O2
vi. Balanced! But look at all the coefficients. They have a common
factor, in this case 2. If you can simplify ALL the coefficients, do
so! In this case, divide ALL coefficients by 2.
SO3 
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SO2 + O2 (the 1 does NOT need to be given)
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Balancing an Equation example #4:
Ca + HNO3  Ca(NO3)2 + H2
i. First thing…treat the ENTIRE polyatomic ion (NO3) as one
element. That means any coefficient affects ALL of the polyatomic
ion atoms. (It may be easier to place any polyatomic ions in
parentheses during balancing.)
Ca + H(NO3)  Ca(NO3)2 + H2
 For now, there is one Ca on both sides of the arrow.
ii. There is one H on the left, and two H on the right. So, (1 x 2 = 2),
so there should be 2 H’s on both sides. Place a coefficient of in
front of HNO3:
Ca +
H(NO3)  Ca(NO3)2 + H2
 That gives two (NO3) polyatomic ions on both sides, one Ca
on both sides, and two H’s on both sides.
iii. Balanced!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Balancing an Equation example #5:
FeCl3 + Pb(NO3)2  Fe(NO3)3 + PbCl2
i. Whoa! There is a LOT here to work with! There are FOUR things
to balance.
ii. Take one species at a time; be patient and do one action at a time.
iii. Start simple; there is one Fe on each side; let that be…for now.
iv. Look at the lead; again, there is only one Pb on each side, so OK.
v. There are three Cl atoms on the reactant side, and two Cl atoms on
the product side. Multiply (2 x 3 = 6), so we need six Cl’s on either
side.
vi. Look at the reactants; divide 6 by the three Cl’s we have on the left,
getting 2, so our coefficient for the reactant FeCl3 will be .
FeCl3 + Pb(NO3)2  Fe(NO3)3 + PbCl2
vii. Look at the products; divide 6 by the two Cl’s we have on the left,
getting 3, so our coefficient for the product PbCl2 will be .
2 FeCl3 + Pb(NO3)2  Fe(NO3)3 +
PbCl2
viii. Now let’s work with the nitrate polyatomic ion. There are two
NO3’s in the reactants and three NO3’s in the products. Again, it is
(2 x 3 = 6), so we’ll need six nitrate ions on either side of the arrow.
ix. Place a coefficient of for the reactant with nitrate, and a coefficient
of
for the product that contains the nitrate ion.
2 FeCl3 +
Pb(NO3)2 
Fe(NO3)3 + 3 PbCl2
x. Hey! Look at the metals! Fe and Pb BOTH have the same amounts
on either side! Note that this doesn’t often happen, but when it
does, sweet!
xi. Balanced!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Writing Equations
Topic:
Objective: How do we write equations when only given names?
Writing Equations when given the names of the formulas:
 If you are given the names of the compounds used in a chemical
reaction rather than their formulas, use can use the same rules for
writing formulas that you used in Unit 7b.
 After writing the formulas and setting up the basic equation, balance
the equation normally.
Given a reaction as names, write the formulas:
1. Magnesium metal + lead (IV) nitrate  magnesium nitrate + lead metal
Mg(s)

+ Pb(NO3)4(aq)
Mg(NO3)2(aq)
+
Pb(s)
Mg(NO3)2(aq) +
Pb(s)
i. Now balance the raw equation:
Mg(s)

+ Pb(NO3)4(aq)
ii. THIS one was simple!
2. Copper metal + Oxygen gas  copper (I) oxide
Cu(s)
+
O2(g)

Cu2O(s)
i. Now balance the raw equation:
Cu(s)
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+
O2(g)

Cu2O(s)
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
3. Lead (II) nitrate + sodium phosphate  lead (II) phosphate + sodium nitrate
Pb(NO3)2(aq) +
Na3PO4(aq)

Pb3(PO4)2(s)
+ NaNO3(aq)
i. Now balance the raw equation:
Pb(NO3)2(aq) +
Na3PO4(aq) 
Pb3(PO4)2(s) +
NaNO3(aq)
Hey! With a little practice, this is easy and fun!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Missing Mass in Equations
Objective: The amount of matter in reactants equals that in products!
The mass on the reactants (left) side of the arrow and the mass on the
products (right) side of the arrow MUST equal each other as the Law of
Conservation of Mass states that mass may not be created or destroyed
in any chemical reaction or physical change.
Missing Mass examples:
1. Given that 35.0 g of N2(g) react with an unknown amount of H2(g) to
produce 42.5 g of NH3(g), how many grams of H2(g) were used?
i. 35.0 g of N2(g) + X g of H2(g) = 42.5 of NH3(g), therefore you first
solve for 42.5 g– 35.0 g = 7.5 g, so 7.5 g of H2(g) was used making
42.5 of NH3(g).
2. How many grams of aluminum are formed when 45.0 g of Al2O3(s)
are decomposed into Al(s) and 21.1 g of O2(g)?
i. 45.0 g of Al2O3 = 21.1 g of O2(g) + X g of Al(s), therefore you first
solve for 45.0 g – 21.1 g = 23.9 g, so 23.9 g of Al(s) is formed.
3. How many grams of FeS(s) must decompose to form 33.0 g of Fe(s)
and 19.0 g of S(s)?
i. 33.0 g of Fe (s) + 19.0 g of S(s) = X g of FeS(s), therefore you first
solve for X = 52.0 g, so 52.0 g of FeS(s) is decomposed.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Practice Regents Questions (ungraded):
1. During all chemical reactions, mass, energy, and charge are
a) Formed
c) Absorbed
b) Released
d) Conserved
2. Which of the following equations show conservation of mass?
a) H2 + O2  H2O
c) 2 H2 + O2  2 H2O
b) H2 + O2  2 H2O
d) 2 H2 + 2 O2  2 H2O
3. Which of the following equations show a conservation of mass?
a) H2O  H2 + O2
c) Na + Cl2  NaCl
d) Al + Br2  AlBr3
b) PCl5  PCl3 + Cl2
4. Which of the following chemical equations is correctly balanced?
c) N2(g) + H2(g)  NH3(g)
a) 2 KCl(s)  2 K(s) + Cl2(g)
b) H2(g) + O2(g)  H2O(g)
d) 2 NaCl(s)  Na(s) + Cl2(g)
Given the following unbalanced equation:
_1_ Fe2O3(s) + _3_ CO(g)  _2_ Fe(s) + _3_ CO2(g)
5. When the equation is correctly balanced using the smallest whole-number
coefficients, what will be the coefficient of CO(g)?
a) 1
b) 2
d) 4
c) 3
Given the following unbalanced equation:
_1_ Mg(ClO3)2(s)  _1_ MgCl2(s) + _3_ O2(g)
6. What is the coefficient of O2(g) when the equation is balanced correctly using the
smallest whole number coefficients?
a) 1
b) 2
d) 4
c) 3
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Student name: _________Key__________ Class Period: _3, 5, & 10_
Please carefully remove this page from your packet to hand in.
Chemical Equations homework
What do each of the following mean in a chemical equation? 1 pt. ea.
(s)
_solid_
(l)
_liquid_
(g)
__gas__
(aq)
_aqueous_
Balance ALL the following equations/reactions by placing a coefficient in front of
the formulas. Do NOT change any subscript or element of each formula of a
compound. You don’t need to put (1) as a coefficient, but may do so if it is
helpful. 1 pt. ea.
Balance the reaction (Fill in the coefficients) 1 pt. ea.
_1_ C(s) + _2_ H2(g)  _1_ CH4(g)
_2_ AgNO3(aq) + _1_ Cu(s)  _2_ Ag(s) + _1_ Cu(NO3)2(aq)
_1_ N2(g) + _2_ O2(g)  _1_ N2O4(g)
_4_ H2O2(l)  _4_ H2O(l) + _2_ O2(g)
simplify coefficients to _2_  _2_ + _1_
_1_ Ca(OH)2(aq) + _2_ HCl(aq)  _1_ CaCl2(aq) + _2_ H2O(l)
Write the formulas for the following compounds. 1 pt. ea.
Compound name
Compound formula
Compound name
Compound formula
Sodium oxide
Na2O
Lead (II) carbonate
PbCO3
Potassium sulfate
K2SO4
Lead (IV) carbonate
Pb(CO3)2
Iron (II) chloride
FeCl2
Zinc phosphate
Zn3(PO4)2
Iron (III) chloride
FeCl3
Zinc phosphide
Zn3P2
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Using the given names, write the formulas and balance the reactions. 1 pt. ea.
Iron + nitrogen gas  iron (II) nitride
_3_ Fe(s) + _1_ N2(g)  _1_ Fe3N2(s)
Lead + copper (II) nitrate  lead (II) nitrate + copper
_1_ Pb(s) + _1_ Cu(NO3)2(aq)  _1_ Pb(NO3)2(aq) + _1_ Cu(s)
Calcium + water  calcium hydroxide + hydrogen gas
_1_ Ca(s) + _2_ H2O(l)  _1_ Ca(OH)2(aq) + _1_ H2(g)
Potassium phosphate + iron (II) nitrate  potassium nitrate + iron (II) phosphate
_2_ K3PO4(aq) + _3_ Fe(NO3)2(aq)  _6_ KNO3(aq) + _1_ Fe3(PO4)2(s)
Find the Missing Mass for each of the problems below. 1 pt. ea.
If 15.0 g of nitrogen gas reacts with 3.2 g of hydrogen gas, how many grams of
ammonia gas will be formed?
_1_ N2(g) + _3_ H2(g)  _2_ NH3(g)
15.0 g of N2(g) + 3.2 g of H2(g) = 18.2 g of N & H  18.2 g of NH3(g)
If 37.0 g of zinc(s) metal reacts with hydrochloric acid (HCl(aq)) to form 77.2 g of
zinc chloride(aq) and 1.1 g of hydrogen gas, how many grams of HCl were reacted?
_1_ Zn(s) + _2_ HCl(aq)  _1_ ZnCl2(aq) + _1_ H2(g)
37.0 g of Zn(s) + X g of HCl = 77.2 g of ZnCl2(aq) + 1.1 g of H2(g)
(-37.0)
X g of HCl = 77.2 + 1.1 - (37.0) = 41.3 g of HCl
How many grams of solid aluminum oxide must decompose to form 112.0 g of
solid aluminum metal and 99.6 g of oxygen gas?
_2_ Al2O3(s)  _4_ Al(s) + _3_ O2(g)
X g of Al2O3(s) = 112.0 g of Al(s) + 99.6 g of O2(g)
211.6 g of Al2O3(s) = 112.0 + 99.6
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Driving Force and Entropy
Objective: What ‘drives’ chemical reactions to occur?
Driving Force:
In nature, changes that require the least amount of energy will be the
changes that occur. Think of a rock; if you let go, it falls. Gravity is the
driving force behind the rock falling; no energy is required on the rock’s
part. The same premise is behind chemical reactions. Reactions will
occur in the direction that requires the least energy to occur.
Entropy as a Measure of Disorder:
Perhaps the best way to understand entropy as a driving force in nature
is to conduct a simple experiment with a new deck of cards. Open the
deck, and remove the jokers. The top card will be the ace of spades,
followed by the two, three, and four of spades, and so on. Now shuffle
the deck, and note that the deck becomes more disordered. The more
often the deck is shuffled, the more disordered it becomes. What makes
a deck of cards become more disordered when shuffled? The force
behind this increased disarray is entropy. The more disorder there is
the more entropy is in the system. If you picked up all the cards and put
them back into the box without sorting, they would be in disorder.
Sorting all the cards requires energy to be input. Reversing entropy
requires greater energy than it did to originally occur.
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Lecture Key
Unit 8: Reactions-Key
Topic:
Regents Chemistry ’14-‘15
Mr. Murdoch
Redox Reactions
Objective: How
do we make
the chemicals of everyday use?
Determining
Molecular
Polarity:
Redox reactions:
 Redox reactions are driven by the loss of electrons (oxidation) and the
gain of electrons (reduction).
 In a redox reaction, one species (an element) gains electrons and the
other species loses electrons. Any ions of elements that are NOT
involved (no loss/no gain of e-) are called “spectator ions”.
 Redox (Reduction-Oxidation): A reaction prompted by an exchange
in electrons between two elements, resulting in a change of the
oxidation numbers of both elements.
i. The more electronegative element (usually a nonmetal) gains the
electron(s), resulting in a decrease in oxidation number (becoming
more negative). This is reduction.
ii. The less electronegative element (usually a metal) loses the
electron(s), resulting in an increase in oxidation number (becoming
less negative). This is oxidation.
Watch YouTube Bozeman Science RedOx Reactions video (11:40)
https://www.youtube.com/watch?v=RX6rh-eeflM
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Lecture Key
Unit 8: Reactions-Key
Topic:
Regents Chemistry ’14-‘15
Mr. Murdoch
Synthesis Reaction
Objective: How do we make a new compound with a redox reaction?
Synthesis Redox Reactions:
*Note: When an element is shown as
X
0
, it means the charge is neutral.
 Synthesis is the creation of something new from separate parts.
 In a synthesis redox reaction, two elements (reactants) combine to
form a new compound (product).
 This is the easiest way to make a new compound, and is often used in
industrial processes.
i. A + B  AB
ii. A0 + B0  A+B- (A0 gives its electron to B0; A0 is oxidized, B0 is
reduced)
 Sample synthesis reaction (Rx):
i. 2 K + Br2  2 KBr
Rx w/ charges: 2 K0 + (Br0 + Br0)  2 K+ 2 Br-
ii. K0 goes from 0 to +1, so K0 is oxidized. K0 gives electrons to Br0.
Br0 goes from 0 to -1, so Br0 is reduced.
Watch YouTube Swarthout Synthesis reaction video (6:49)
https://www.youtube.com/watch?v=MhlWTZwDHM8
Watch YouTube Swarthout Decomposition reaction video (8:13)
https://www.youtube.com/watch?v=1ocQhkHw_MM
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Decomposition Reaction
Objective: How do we break down a compound into its elements?
Decomposition Redox Reactions:
 Decomposition is taking a compound and breaking it down to its
individual elements it was created from.
 Decomposition redox reactions are rare in nature. This is due to the
fact that this requires more energy to break down compounds than
it took to form the compound. Water does not spontaneously break
down into hydrogen gas and oxygen; table salt does not
spontaneously break down into sodium metal and chlorine gas. Both
decompositions require a lot of energy.
 Decomposition reactions are used to get pure highly reactive elements
(sodium, oxygen, chlorine, etc.) out of the naturally occurring
compounds where they are found.
 Decomposition is often done using electrolytic decomposition, or
using electricity to break down a compound.
i. AB  A + B
ii. A+B-  A0 + B0 (B- gives its electron to A+, so A0 is reduced and
B0 is oxidized.)
 Sample decomposition reaction (Rx):
i. 2 AgCl  2 Ag + Cl2
Rx w/ charges: 2 Ag+1Cl-1  2 Ag0 + Cl20
ii. Ag+1 went from +1 to 0, so Ag+1 is reduced. Ag+1 gains electrons
from Cl-1; Cl-1 goes from -1 to 0 and is oxidized.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Single Replacement Reaction
Objective: How do we work with a single element and compound?
Single Replacement Reaction:
 Single replacement reactions have one single element and one type of
compound on either side of the reaction arrow.
o There are two possibilities of Single Replacement Reactions:
i. A metal plus a compound;
ii. A nonmetal plus a compound.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Single Replacement - Metal
Objective: How do we replace positive ions within a compound?
Single Replacement Reaction with a Metal:
1. Metal plus a compound single replacement.
i. The metal replaces the positive ion in the compound.
ii. This reaction may be used in batteries to create an electrical
current.
iii. This reaction may also be used to remove a less reactive element
from a compound.
iv. Adding a more reactive metal to the compound containing the less
reactive metal will “drive” the less reactive metal out, allowing
the more reactive metal to become part of the compound. The less
reactive metal will now be in its pure form.
v. This process is sometimes used in extraction of metal ore.
a. A + BC  AC + B
b. A0 + B+C-  B+A- + C0 (C- gives its electrons to A0, so C- is
oxidized and A0 is reduced.) B+ does not have a change in
charge, so B+ is a spectator ion.
 Sample Single Replacement Reaction with a Metal:
i. Reaction: Zn + Cu(NO3)2  Zn(NO3)2 + Cu
ii. Rx w/ charges: Zn0 + Cu+2 (NO3)2-1  Zn+2(NO3)2-1 + Cu0
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
iii. Zn0 goes from 0 to +2, so it is oxidized. Zn0 loses electrons to
Cu+2, so Cu+2 goes from +2 to 0 and is reduced. NO3-1 does not
have a change in charge, so NO3-1 is a spectator ion.
Watch YouTube Swarthout Single Replacement reaction video (9:27)
https://www.youtube.com/watch?v=qhmTXdOKTBo
Website upload 2015
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Single Replacement-Nonmetal
Objective: How do we replace negative ions within a compound?
Single Replacement Reaction with a Nonmetal:
1. Nonmetal plus a compound single replacement.
i. The nonmetal replaces the negative ion in the compound.
ii. This is a rare type of reaction, and is a more costly method to
produce a pure nonmetal than using electrolytic decomposition.
a. A + BC  BA + C
b. A0 + B+C-  B+A- + C0 (C- gives its electrons to A0, so C- is
oxidized and A0 is reduced.) B+ does not undergo a change in
charge, so B+ is a spectator ion.
 Sample Replacement Reaction with a Nonmetal:
i. Reaction: F2 + ZnCl2  ZnF2 + Cl2
ii. Reaction w/ charges: F20 + Zn+2Cl2-1  Zn+2F2-1 + Cl20
iii. F0 goes from 0 to -1, so F0 is reduced. F0 gains electrons from the
Cl-1, which goes from -1 to 0 and Cl-1 is oxidized. Zn+2 will not
have a change in charge, so Zn+2 is a spectator ion.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Identifying Reaction Type
Objective: How do identify the type of redox reaction?
Identifying the type of Reaction given:
1. 2 K + Zn(NO3)2  2 KNO3 + Zn
 This reactions starts with an element and a compound and ends
with an element and a compound; it must be a single replacement
reaction.
2. 2 Fe + 3 Cl2  2 FeCl3
 This reaction starts with elements and ends with a compound; it
must be a synthesis reaction.
3. 2 NaNO2  2 Na + N2 + 2 O2
 This reaction starts with a compound and ends with elements; it
must be a decomposition reaction.
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Page 31 of 57
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Determining Species Charge
Objective: How we determine the charges of the reaction species?
Determining the charge of a reaction species:
i. Elements have NO charge; elements are neutral species.
ii. The charge for ions of elements (charged atoms) may be found
using the periodic table. If the metal (cation) has multiple charges
listed, use the nonmetal (anion) to determine the cation charge.
iii. Polyatomic ions have their charges listed in Reference Table E.
 Examples:
1. Al2(SO4)3 ̶ Al is listed as forming only a +3 cation on the periodic
table, and the sulfate polyatomic ion is found as -2 on Table E as
SO4-2. Each of the three polyatomic ion has a -2 charge, for a total
anion charge of -6, so the cations must equal a total +6 charge,
forcing the two Al atoms to form Al+3 ions:
Al2+3(SO4)3-2
̶ Fe has two charges listed (+2 & +3) on the periodic
table. The phosphate polyatomic ion is found as -3 on Reference
Table E as PO4-3. As there are two phosphate polyatomic ions,
each with a -3 charge, the total negative charge is -6. Therefore the
total positive charge MUST be +6. As there are 3 Fe atoms, each
Fe atom needs to have a +2 charge to create the total +6 charge:
2. Fe3(PO4)2
Fe3+2(PO4)2-3
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
3. 4 Ag0 + O20  2 Ag2+1O-2 ̶ Note that the elements (Ag & O) are
by themselves and each have a charge of 0. O2 is a diatomic
molecule; as such it is neutral (charge of 0).
4. 6 K0 + Al2+3(SO4)3-2  3 K2+1SO4-2 + 2 Al0 ̶ Note that the SUM of
the ion charges in each of the compounds totals ZERO.
5. 2 Fe+2O-2  2 Fe0 + O20
6. 2 Na0 + Br20  2 Na+1Br-1
7. Ca0 + 2 H+1Cl-1  Ca+2Cl2-1 + H20
8. 2 Pb+2O-2  2 Pb0 + O20
9. 2 Al0 + 3 Ag2+1CO3-2  Al2+3(CO3)3-2 + 6 Ag0
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Determining Species Type
Objective: How do we find the oxidized, reduced, or spectator ion?
Determining Species Type in a Reaction:
i. The species in a reaction that is OXIDIZED lost electrons, and the
oxidation number has become more POSITIVE.
ii. The species in a reaction that is REDUCED gained electrons, and
the oxidation number has become more NEGATIVE.
iii. The species that is the SPECTATOR ION neither gains nor loses
electrons, and its charge remains the SAME.
Examples:
OX = Oxidized species
RD = Reduced species
SI = Spectator ion
1. 4 Ag0 + O20  2 Ag2+1O-2
OX (charge ↑ +): Ag0
RD (charge ↑ -): O20
SI: none
2. 6 K0 + Al2+3(SO4)3-2  3 K2+1SO4-2 + 2 Al0
OX (charge ↑ +): K0
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RD (charge ↑ -): Al+3
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SI: SO4-2
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
3. 2 Fe+2O-2  2 Fe0 + O20
OX (charge ↑ +): O-2
RD (charge ↑ -): Fe+2
SI: none
RD (charge ↑ -): Br20
SI: none
4. 2 Na0 + Br20  2 Na+1Br-1
OX (charge ↑ +): Na0
5. Ca0 + 2 H+1Cl-1  Ca+2Cl2-1 + H20
OX (charge ↑ +): Ca0
RD (charge ↑ -): H+1
SI: Cl-1
RD (charge ↑ -): Pb+2
SI: none
6. 2 Pb+2O-2  2 Pb0 + O20
OX (charge ↑ +): O-2
7. 2 Al0 + 3 Ag2+1CO3-2  Al2+3(CO3)3-2 + 6 Ag0
OX (charge ↑ +): Al0
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RD (charge ↑ -): Ag+1
Page 35 of 57
SI: CO3-2
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Finding a Missing Species
Objective: How do we find which part of a reaction is missing?
Determining the missing species in a reaction:
If the missing species is an ELEMENT, write its symbol. If the missing
species is diatomic (Br2, I2, N2, Cl2, H2, O2, F2), write it as a diatomic
molecule. This will balance the equation.
If the missing species is a COMPOUND, write the negative ion first and
the positive ion second. Write the formula in such a way that you have
the equal numbers of atoms of each element on both sides of the
balanced reaction.
Examples:
1. 4 Ag + O2  2 _ Ag2O _
(missing species is Ag2O)
i. There are now 4 Ag and 2 O on each side of the reaction arrow.
2. 6 K + Al2(SO4)3  3 _ K2SO4_ + 2 Al
(missing species is K2SO4)
i. There are 6 K’s and 3 SO4’s missing, so with the coefficient of 3,
the new compound contains 2 K’s (3 x 2) and 1 SO4 (3 x 1).
ii. There are two Al’s on both sides…OK!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
3. 2 FeO  2 Fe + ___ O2___
(missing species is O2)
i. O2 is diatomic, so there are now 2 O’s on each side.
ii. There are two Fe’s on both sides…OK!
4. 2 Al + 3 _ Ag2CO3_  Al2(CO3)3 + 6 Ag
(Ag2CO3 is missing)
i. There are 6 Ag’s and 3 CO3’s missing. Since the coefficient is 3,
the compound contains 2 Ag’s (3 x 2) and 1 CO3 (3 x 1).
ii. There are two Al’s on both sides…OK!
5. 2 Na + Br2  2 _NaBr_
6. Ca + 2 __HCl__

CaCl2 + H2
7. 2 PbO  2 ___Pb___ + O2
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(missing species is NaBr)
(missing species is HCl)
(missing species is Pb)
Page 37 of 57
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Regents Practice Problems (ungraded):
Given the balanced equation:
2 KClO3  2 KCl + 3 O2
1. Which type of reaction is represented by the above equation?
a) Synthesis
c) Single Replacement
d) Nuclear Transmutation
b) Decomposition
Given the balanced equation representing the reaction:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
2. Which type of chemical reaction is represented by the above equation?
c) Single Replacement
a) Synthesis
b) Decomposition
d) Nuclear Transmutation
3. Which of the balanced equations below represent a chemical change?
a) H2O(l)  H2O(s) + energy
b) H2O(g)  H2O(l) + energy
c) H2O(l) + energy  H2O(g)
d) 2 H2O(l) + energy  2 H2(g) + O2(g)
4. Which of the following lists includes three types of chemical reactions?
a) Condensation, solidification, and synthesis
b) Decomposition, solidification, and sublimation
c) Decomposition, single replacement, and synthesis
d) Condensation, single replacement, and sublimation
Given the balanced reaction:
Mg(s) + 2 AgNO3(aq)  Mg(NO3)2(aq) + 2 Ag(s)
5. Which type of reaction is represented above?
a) Synthesis
c) Single Replacement
b) Decomposition
d) Nuclear Transmutation
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Student name: _________Key__________
Class Period: _3, 5, & 10_
Please carefully remove this page from your packet to hand in.
Oxidation and reduction Reactions homework
*Note: Make sure to include ALL charges when writing a species!
Write Al+3, not Al
Write Na0, not Na
Write Cl-1, not Cl
Multiple choice questions: circle the correct answer choice. 1 pt. ea.
1. In the reaction 2 Na + 2 HNO3  2 NaNO3 + H2, which species is the spectator
ion?
a) Na0
b) H+1
d) Na+1
c) NO3-1
2. In the reaction Zn + 2 HNO3  ________ + H2, the missing species is:
c) ZnNO3
a) Zn(NO3)2
b) (NO3)2Zn
d) NO3Zn
3. In the reaction 2 ________  2 Pb + O2, the missing species is:
b) OPb
c) PbO2
a) PbO
d) O2Pb
Identify the type of reaction (synth, decomp, or sngl rplcmnt) below. 1 pt. ea.
Given Reaction
KBr + Na  NaBr + K
Reaction type How can you tell the type?
sngle rplcmnt Cmpd + Elmt  Cmpd + Elmt
2 LiNO3 + Ca  Ca(NO3)2 + 2 Li
sngl replcmnt Cmpd + Elmt  Cmpd + Elmt
2 Fe + 3 Cl2  2 FeCl3
synthesis
Elmt + Elmt  Cmpd
2 Li2O  4 Li + O2
decomp
Cmpd  Elmt + Elmt
Write the charges of each of the species in the reactions below, then identify which
species are oxidized, reduced, or are spectator ions (if any are present). 1 pt. ea.
Ag2S  2 Ag + S
Oxidized: ___S-2___
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2 Ag+1 S-2  2 Ag0 + S0
Reduced: ___Ag+1___
Spectator ion: __none__
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
K+1 Br-1 + Na0  Na+1 Br-1 + K0
KBr + Na  NaBr + K
Oxidized: ___Na0___
Reduced: ___K+1___
2 LiNO3 + Ca  Ca(NO3)2 + 2 Li
Oxidized: ____Ca0____
2 {Li+1 NO3-1} + Ca0  Ca+2 (NO3)2-1 + 2 Li0
Reduced: ___Li+1___
Spectator ion: ___NO3-1___
2 Fe0 + 3 Cl20  2 {Fe+3 Cl3-1}
2 Fe + 3 Cl2  2 FeCl3
Oxidized: ____Fe0____
Spectator ion: ___Br-1___
Reduced: ___Cl20___
Spectator ion: ___none___
Complete the following reactions by writing the appropriate formula and balancing
the equations. 1 pt. ea.
2 Na + Br2  2 __NaBr__
3 Zn + 2 Fe(NO3)3  3 __Zn(NO3)2__ + 2 Fe
2 BaO  2 Ba + __O2__
2 Sc + 3 CuSO4  Sc2(SO4)3 + 3 ____Cu____
2 NO2  ____N2____ + 2 O2
2 ____Li____ + S  Li2S
Cu + 2 ___KNO2___  Cu(NO2)2 + 2 K
Mg + ____Cl2____  MgCl2
2 ___Li3N___  6 Li + N2
2 Na + ___CaCO3___  Na2CO3 + Ca
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Solubility & Ionic Compounds
Objective: How do ionic substances interact in aqueous solutions?
Aqueous Ionic Compounds:
1. Soluble Ionic Compounds:
i. Ionic compounds generally ionize in water. When a soluble ionic
compound is placed into water, the interaction between the strong
polar bonds of the water molecule and the weaker bonds of the
ionic compound rips the ionic compound into its component ions.
ii. The more electronegative (δ-) oxygen end of the water molecule
attracts and attaches to the more positive ion. The less
electronegative (δ+) hydrogen end of the water molecule attracts and
attaches to the more negative ion. The liquid water molecules,
continually moving, tear the crystal ions apart and keep the ions
apart in solution.
2. Insoluble Ionic Compounds:
i. Insoluble compounds have ionic structures that remain together
when in water. This is because the attractions between the ions are
too strong for water molecules to tear them apart. Insoluble ions
attach to other insoluble ions, and the particles form a cloudy
suspension. The suspension will eventually form solid crystals,
called a precipitate, which settles due to gravity.
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Lecture Key
Unit 8: Reactions-Key
Topic:
Regents Chemistry ’14-‘15
Mr. Murdoch
Determining Solubility
Objective: How do we find which ionic compounds are soluble?
Which Ionic Compounds are Soluble (or Insoluble) in water?
 To determine if an ionic compound is soluble or insoluble in water,
use Reference Table F, Solubility Guidelines for Aqueous Solutions.
 Note that there are two columns, Ions That Form Soluble
Compounds, and Ions That Form Insoluble Compounds in
Reference Table F. Note that to the right of each column is the
Exceptions column for either Soluble or Insoluble ions. The
Exceptions column gives examples when the general Soluble or
Insoluble guideline is NOT applicable for that type of ion. Be careful
when reading Table F; it can cause headaches!
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Using Table F for Solubility
Objective: How do we interpret Reference Table F?
Use Reference Table F, Solubility Guidelines for Aqueous Solutions, to
determine the reason why a compound is soluble or insoluble in water.
Ionic
Soluble or
Compound Insoluble?
Reason why is Soluble or Insoluble.
Li2CO3
Soluble
Li+1 is a Group 1 ion, and ALL Group 1 ions are soluble
(blank exceptions)
Pb(NO3)2
Soluble
NO3-1 (nitrate) is given as soluble (blank exceptions)
ZnCl2
Soluble
Cl-1 is a halide, given as soluble, with few exceptions
(Zn+2 is NOT an exception)
BaSO4
Insoluble
SO4-2 (sulfate) is given as soluble, with few exceptions
(Ba+2 IS an exception)
Ca(HCO3)2 Soluble
HCO3-1 (hydrogen carbonate) is given as soluble (blank
exceptions)
CuCO3
Insoluble
CO3-2 (carbonate) is given as insoluble, with few
exceptions (Cu+2 is NOT an exception, as Cu+2 is NOT a
Group 1 ion)
Insoluble
CrO4-2 (chromate) is given as being insoluble, with few
exceptions (Pb+2 is NOT and exception, as Pb+2 is NOT a
Group 1 ion)
Soluble
PO4-3 (phosphate) is insoluble EXCEPT with Group 1 or
NH4+1 (Na+1 is a Group 1 ion, so Na+1 IS an exception)
Pb(CrO4)2
Na3PO4
NH4OH
Soluble
Mg(OH)2
Insoluble
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NH4+1 (ammonium) is given as always soluble (blank
exceptions)
OH-1 (hydroxide) is given as insoluble, with few
exceptions (Mg+2 is NOT an exception, as Mg+2 is NOT a
Group 1 ion)
Page 43 of 57
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Double Replacement Reaction
Objective: What happens in a double replacement reaction?
Double Replacement Reactions:
 In a double replacement reaction the positive ion of one ionic
compound swaps places with the negative ion from the other ionic
compound. The double replacement reaction is carried out within
aqueous solutions of each reactant ionic compound so that the ions
may more easily mix and react.
+
+
 General Double Replacement Formula:
A+B-(aq) + C+D-(aq)  A+D-(s) + C+B-(aq)
Watch YouTube Swarthout Double Replacement reaction video (10:49)
https://www.youtube.com/watch?v=7hVKb4ROjZw
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Page 44 of 57
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Aqueous solution AB (composed of separated A+ and B- ions) is mixed
into aqueous solution CD (composed of separated C+ and D- ions). As
they are mixed, the A+ ions of the first solution seek out and tightly bond
with the D- ions from the second solution. These A+ and D- ions bond so
tightly that that the polar water molecules cannot strip them apart. The
resulting ionic compound AD is formed of small crystals that turn the
combined solution cloudy. As the AD crystals clump together and gain
mass, they will settle out due to gravity. The settled solid AD is the
precipitate of the reaction.
Examples of Double Replacement Reactions:
1. K2CrO4(aq) + Ba(NO3)2(aq)  2 KNO3(aq) + BaCrO4(s)
i. Look for the products (K+ and NO3-) and (Ba+2 and CrO4-2) in
Reference Table F. K+1 is a Group 1 ion; K+ is always soluble.
NO3- is given as always soluble as well. The KNO3 will remain as
aqueous (in solution with water). The CrO4-2 is listed as insoluble,
and Ba+2 is NOT one of the exceptions for CrO4-2. Therefore the
BaCrO4 is the solid precipitate that forms from this reaction.
2. Na2CO3(aq) + CaCl2(aq)  2 NaCl(aq) + CaCO3(s)
i. Look for the products (Na+ and Cl-) and (Ca+2 and CO3-2) in Table F.
Na+1 is another Group 1 ion, and always soluble. Cl-1 is usually
soluble, and Na+1 is not an exception here. The NaCl will remain as
aqueous. The CO3-2 is given as insoluble, and Ca+2 is NOT an
exception for CO3-2. Therefore the CaCO3 is the solid precipitate
that forms from this reaction.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Double Replacement Reaction
Objective: How can double replacement reactions be completed?
Completing a Double Replacement Reaction:
How may we complete a Double Replacement Reaction that contains
missing compounds? The same way we did earlier with single redox
reactions.
i. For the missing compound, write the negative ion first and the positive
ion second. Write the formula in such a way that you have the equal
numbers of atoms of each element on both sides of the balanced
reaction.
ii. Once you have the balanced reaction, use Reference Table F to
identify the precipitate in the reaction.
*NOTE: Not all double replacement reactions form precipitates!
 Examples of Completing a Double Replacement Reaction:
I. KCl(aq) + AgNO3(aq)  KNO3(_____) + _________(_____)
a) The missing compound will contain Ag+1 and Cl-1. Since no
coefficient was given, the formula will be AgCl. Using
Reference Table F, KNO3 is soluble (will be aqueous), and the
AgCl(s) will be the insoluble precipitate.
KCl(aq) + AgNO3(aq)  KNO3(__aq__) + __AgCl__(__s__)
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Lecture Key
Unit 8: Reactions-Key
II.
Regents Chemistry ’14-‘15
Mr. Murdoch
2 Na3PO4(aq) + 3 Mg(NO3)2  6 NaNO3(______) + ___________(_____)
a) The missing compound contains Mg and PO4. The reactants
contain 2 PO4’s and 3 Mg’s. There is no coefficient given for the
missing product, so the formula is Mg3(PO4)2. Using Table F,
NaNO3 is soluble and will remain in aqueous solution, and
Mg3(PO4)2(s) will be the insoluble precipitate.
2 Na3PO4(aq) + 3 Mg(NO3)2  6 NaNO3(__aq__) + _Mg3(PO4)2_(__s__)
The Double Replacement reactions shown in examples I and II both
formed solid precipitates. While most double replacement reactions we
will work with do form solid precipitates, two other double replacement
reactions do NOT form precipitates. They are neutralization and gasforming reactions.
 If solution AB is an acid (consisting of AH+ and BNnM- aqueous ions)
and solution CD is a base (consisting of CM+ and DOH- ions). When a
double displacement reaction occurs, the cations and anions switch
partners, resulting in the formation of water and a new ionic
compound (or salt), which is usually soluble.
 Example of a neutralization reaction:
a) Sulfuric acid(aq) + lithium hydroxide(aq)  water(l) + lithium sulfate(aq)
b) Equation: H2SO4(aq) + 2 LiOH(aq)  Li2SO4(aq) + 2 H2O(l)
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Identifying Spectator Ions
Objective: How do we find the aqueous ions that remain unchanged?
Spectator Ion Determination:
 In a Double Replacement Reaction the ions that remain dissolved in
the solution (that remain aqueous) are called Spectator Ions.
 We may identify the Spectator Ions by determining which ions
remain in aqueous solution on BOTH sides of the equation.
Mg(NO3)2(aq) + Na2CO3(aq)  MgCO3(s) + 2 NaNO3(aq)
 Compare the reactants (right side) with the products (left side):
i. Mg+2 is found in the product precipitate; it underwent a replacement.
ii. NO3-1 is found in aqueous compounds on BOTH sides; it was
unchanged.
iii. Na+1 is found in aqueous compounds on BOTH sides; it was
unchanged.
iv. CO3-2 is found in the product precipitate; it underwent a
replacement.
v. As the NO3-1 and Na+1 are the only ions that started in aqueous
solution and finished in aqueous solutions, they are the spectator
(unchanged) ions in this precipitate reaction.
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Student name: __________Key__________
Class Period: _3, 5, & 10_
Please carefully remove this page from your packet to hand in.
Double Replacement Homework
Using Reference Table F, determine if the following compounds are soluble or
insoluble and give your reason(s). 1 pt. ea.
Na2CO3
Soluble or
Insoluble
S
CaS
I
Sulfide is insoluble with all but Grp 1 & NH4+1
PbBr2
I
Halides (Br) are insoluble with lead (II)
Al(OH)3
I
Hydroxide is insoluble with all but Grp 1, some Grp 2, & NH4+1
(NH4)2CrO4
S
Ammonium is listed as exception to insoluble CrO4-2
Compound
Formula
Reasons for your choice
Sodium is a Group 1 exception for CO3-2
CIRCLE the precipitate (insoluble product) in each double replacement reaction.
Reaction (Circle each precipitate compound in the product) 1 pt. ea.
3 Ca(NO3)2(aq) + 2 K3PO4(aq)  6 KNO3(aq) + Ca3(PO4)2(s)
AgNO3(aq) + NaOH(aq)  NaNO3(aq) + AgOH(s)
Na2SO4(aq) + Ba(ClO3)2(aq)  NaClO3(aq) + BaSO4(s)
(NH4)2CO3(aq) + Ca(C2H3O2)2(aq)  CaCO3(s) + 2 NH4C2H3O2(aq)
2 AgNO3(aq) + Li2CO4(aq)  2 LiNO3(aq) + Ag2CrO4(s)
Write the formula for the missing product (give as (s) or (aq) as well) in these double
replacement reactions. 2 pts ea.
i. CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2 _____NaCl(aq)_____
ii. (NH4)2CO3(aq) + Cu(NO3)2(aq)  CuCO3(s) + 2 ____NH4NO3(aq)____
iii. 2 Li3PO4(aq) + 3 Fe(NO3)2(aq)  6 LiNO3(aq) + ___Fe3(PO4)2(s) ___
iv. Pb(ClO3)2(aq) + 2 KBr(aq)  PbBr2(s) + 2 ____KClO3(aq)___
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
For each of the multiple choice questions below circle the correct answer. 1 pt. ea.
1. Which of the following equations represent a double replacement reaction?
a) 2 H2(g) + O2(g)  2 H2O(l)
b) CaCO3(s)  CaO(s) + CO2(g)
c) KOH(aq) + HCl(aq)  KCl(aq) + H2O(l)  Acid-Base Neutralization Rx
d) Cu(s) + 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
Given the balanced equation:
AgNO3(aq) + NaCl(aq)  NaNO3(aq) + AgCl(s)
2. The reaction above may best be classified as
a) Synthesis
c) Single Replacement
b) Decomposition
d) Double Replacement
3. Which of the following equations represent a double replacement reaction?
a) CaCO3(s)  CaO(aq) + CO2(g)
b) LiOH(aq) +HCl(aq)  LiCl(aq) + H2O(l)  Acid-Base Neutralization Rx
c) CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
d) 2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
Given the balanced reaction:
Ba(NO3)2(aq) + Na2SO4(aq)  2 NaNO3(aq) + BaSO4(s)
4. Which of the following is the name for the precipitate formed?
a) Nitrate oxide
c) Barium nitrate
b) Sodium oxide
d) Barium sulfate
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Stoichiometry of Equations
Objective: How do we make the amount of product needed?
A chemical reaction is basically a recipe used to make something new
from “ingredients”. Just like any other recipe, we can make a smaller or
larger amount by changing the amount of the reactants used.
For chemical reactions the coefficients in front of each species tells you
what the proportional number of moles for each reactant is needed to
make the proportional number of moles of product. The reaction
below is the Haber Process to manufacture ammonia:
N2(g) + 3 H2(g)  2 NH3(g)
The reaction above states that when nitrogen and hydrogen are reacted in
a 1:3 mole ratio, you will produce 2 moles of ammonia. The same 1:3
ratio applies to a small batch of ammonia, or a larger batch. Therefore,
if asked to produce a total of 10 moles of ammonia, you can easily scale
up the “recipe” to make what is asked.
When you divide the “asked” amount (10 moles) by the coefficient for
the reaction (2 moles), you have a factor of 5. This factor of 5 is then
multiplied by the reactants (5 x 1 mole of N2(g)) and (5 x 3 moles of
H2(g)) which will result in the production of 10 moles of NH3(g).
(5 x 1) moles of N2 needed
1 N2(g)
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(5 x 3) moles of H2 needed
+
3 H2(g)

Page 51 of 57
Asked for 10 moles of NH3 (10/2 moles) = 5x
2 NH3(g)
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Topic: Stoichiometry of Equations
Objective: How do we make the amount of product needed?
Question asked: How many moles of product ‘X’ are formed when ‘n’ moles of ‘Y’ react?
How may we use this simple mole ratio relationship to determine what
to use in a chemical reaction? There is a very simple equation to use:
Moles of given x
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛
= Moles of target
Note that for the equation above and for each of the following examples,
the “given” mole ratio is written in strikethrough font, indicating that
this unit cancels out to leave the target.
Stoichiometry Equation examples:
For the reaction N2(g) + 3 H2(g)  2 NH3(g), how many moles of NH3(g) will be formed
when 6.0 moles of N2(g) are completely reacted with excess (more than needed) H2(g)?
Moles of given x
6.0 moles of N2(g) x
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛
2 moles of NH3(g)
1 mole of N2(g)
= Moles of target
= 12. moles of NH3(g)
6.0 moles of N2(g) reacted with excess H2(g) will form 12. moles of NH3(g)
For the reaction N2(g) + 3 H2(g)  2 NH3(g), how many moles of H2(g) will be needed to
completely react with excess (more than needed) N2(g) to form 1000. moles of NH3(g)?
Moles of given x
1000. moles of NH3(g) x
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛
3 moles of H2(g)
2 moles of NH3(g)
= Moles of target
= 1500. moles of H2(g)
1500. moles of H2(g) reacted with excess N2(g) will form 1000. moles of NH3(g)
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
For the reaction 2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g), how many moles of Na(s) are
needed to form 4.0 moles of H2(g)?
Moles of given x
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛
2 moles of Na(s)
1 mole of H2(s)
4.0 moles of H2(g) x
= Moles of target
= 8.0 moles of Na(s)
8.0 moles of Na(s) are reacted with excess H2O(l) and NaOH(aq) to form 4.0 moles of H2(g)
For the reaction 4 Al(s) + 3 O2(g)  2 Al2O3(s), how many moles of Al2O3(s) will form if 6.0
moles of Al(s) are reacted in excess O2(g)?
Moles of given x
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛
2 moles of Al2O3(s)
4 moles of Al(s)
6.0 moles of Al(s) x
= Moles of target
= 3.0 moles of Al2O3(s)
3.0 moles of Al2O3(s) will be formed if 6.0 moles of Al(s) react with excess O2(g)
For the reaction 2 NaCl(aq) + Pb(NO3)2(aq)  2 NaNO3(aq) + PbCl2(s), how may moles of
PbCl2(s) precipitate will form when 5.0 moles of Pb(NO3)2(aq) are reacted in excess with
NaCl(aq)?
Moles of given x
5.0 moles of Pb(NO3)2(aq) x
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛
= Moles of target
1 mole of PbCl2(s)
= 5.0 moles of PbCl2(s)
1 mole of Pb(NO3)2(aq)
5.0 moles of PbCl2(s) will form when 5.0 moles of PbCl2(s) react in excess NaCl(aq)
Watch YouTube Bozeman Science Stoichiometry video (9:45)
https://www.youtube.com/watch?v=LQq203gyftA
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Practice Regents Questions (ungraded):
Given the balanced equation for the reaction:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
1. What is the total number of moles of O2(g) required for the complete combustion
(reaction) of 1.5 moles of C3H8(g)?
a) .30 moles of O2(g)
c) 4.5 moles of O2(g)
b) 1.5 moles of O2(g)
d) 7.5 moles of O2(g)
Given the balanced equation for the reaction:
2 CO(g) + O2(g)  2 CO2(g)
2. What is the mole ratio of CO(g) to CO2(g) in this reaction?
c) 2:1
a) 1:1
b) 1:2
d) 3:2
Given the balanced reaction between the gases propane and oxygen:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
3. Which of the following shows the correct fractional ratio of oxygen to propane?
a)
5 grams of O2
1 gram of C3H8
c)
10 grams of O2
11 grams of C3H8
b)
5 moles of O2
1 mole of C3H8
d)
10 moles of O2
11 moles of C3H8
Given the reaction:
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
4. How many moles of C6H12O6(s) are needed to produce 24 moles of carbon
dioxide gas?
a) 1.0 moles
c) 12. moles
d) 24. moles
b) 4.0 moles
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Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Student name: _________________________
Class Period: _______
Please carefully remove this page from your packet to hand in.
Stoichiometry of Equations homework
For ALL mole conversion problems given here, show ALL of your work,
including showing which units cancel by lining them out.
1. Given the reaction of N2(g) + H2(g)  NH3(g), answer the following. 1 pt. ea.
i. Balance the reaction: __1__ N2(g) + __3__ H2(g)  __2__ NH3(g)
ii. How many moles of N2(g) are needed to form 5.0 moles of NH3(g) reacted in excess H2(g)?
5.0 moles of NH3(g) x
1 mole of N2(g)
2 moles of NH3(g)
= 2.5 moles of N2(g) needed
iii. How many moles of N2(g) are needed to completely react with 10.0 moles of H2(g)?
10.0 moles of H2(g) x
1 mole of N2(g)
3 moles of H2(g)
= 3.33 moles of N2(g) needed
iv. How many moles of NH3(g) should form from 6.0 moles of H2 reacted in excess N2(g)?
6.0 moles of H2(g) x
2 moles of NH3(g)
3 moles of H2(g)
= 4.0 moles of NH3(g) needed
2. Given the reaction of Zn(s) + HBr(aq)  ZnBr2(aq) + H2(g), answer the following. 1 pt. ea.
i. Balance the reaction: __1__ Zn(s) + __2__ HBr(aq)  __1__ ZnBr2(aq) + __1__ H2(g)
ii. How many moles of Zn(s) are needed to form 8.0 moles of ZnBr2(aq) in excess HBr(aq)?
8.0 moles of ZnBr2(aq) x
1 mole of Zn(s)
1 mole of ZnBr2(aq)
= 8.0 moles of Zn(s) needed
iii. How many moles of HBr(aq) are needed to form 4.0 moles of H2(g)?
4.0 moles of H2(g) x
2 moles of HBr(aq)
1 mole of H2(g)
= 8.0 moles of HBr(aq) needed
iv. How many moles of ZnBr2(aq) will form if 5.0 moles of HBr(aq) are reacted in excess Zn(s)?
5.0 moles of HBr(aq) x
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1 moles of ZnBr2(aq)
2 moles of HBr(aq)
Page 55 of 57
= 2.5 moles of ZnBr2(aq) needed
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
3. Given the reaction of Ca(s) + N2(g)  Ca3N2(s), answer the following. 1 pt. ea.
i. Balance the reaction: __3__ Ca(s) + __1__ N2(g)  __1__ Ca3N2(s)
ii. How many moles of Ca(s) are required to form 4.00 moles of Ca3N2(s) in excess N2(g)?
4.00 moles of Ca3N2(s) x
3 moles of Ca(s)
1 mole of Ca3N2(s)
= 12.0 moles of Ca(s) needed
iii. How many moles of Ca3N2(s) will form if 2.50 moles of N2(g) react with excess Ca(s)?
2.50 moles of N2(g) x
1 mole of Ca3N2(s)
1 mole of N2(g)
= 2.50 moles of Ca3N2(s) needed
iv. How many moles of Ca(s) are needed to completely react with 5.00 moles of N2(g)?
5.00 moles of N2(g) x
3 moles of Ca(s)
1 mole of N2(g)
= 15.0 moles of Ca(s) needed
4. Given the reaction of Al(s) + MgCl2(aq)  AlCl3(aq) + Mg(s), answer the following. 1 pt. ea.
i. Balance the equation: __2__ Al(s) + __3__ MgCl2(aq)  __2__ AlCl3(aq) + __3__ Mg(s)
ii. How many moles of Mg(s) will form if 35.0 moles of Al(s) react in excess MgCl2(aq)?
35.0 moles of Al(s) x
3 moles of Mg(s)
2 moles of Al(s)
= 52.5 moles of Mg(s) needed
iii. How many moles of MgCl2(aq) are required to completely react with 15.0 moles of Al(s)?
15.0 moles of Al(s) x
3 moles of MgCl2(aq)
= 22.5 moles of MgCl2(aq) needed
2 moles of Al(s)
iv. How many moles of AlCl3(aq) will be formed if 13.0 moles of Mg(s) are formed?
13.0 moles of Mg(s) x
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2 moles of AlCl3(aq)
3 moles of Mg(s)
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= 8.66 moles of AlCl3(aq) needed
Lecture Key
Unit 8: Reactions-Key
Regents Chemistry ’14-‘15
Mr. Murdoch
Notes page:
Website upload 2015
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Lecture Key