Download Chemical Reactions and Stoichiometry

Chemical Reactions and Stoichiometry
Chapter 8 Balancing Chemical Reactions (Wednesday 1/5/11 P1,3,5 & Thursday 1/6/11 P2,4,6)
I.
II.
III.
Indicators of a Chemical Reaction – evidence of a chemical reaction
a. Evolution of heat and light (simultaneously)
b. Production of a gas (bubbles, odor change)
c. Formation of a precipitate (solid, cloudy)
d. Color change (not introduced by an outside source such as dye or ink)
Characteristics of a Chemical Reaction – the atoms in one or more reactant
rearrange when bonds are broken and/or created to produce one or more new
products with different properties than the reactants had before the reaction.
a. Reactants - on the left– react to one another and/or their environment.
b. Products – on the right of the reaction – produced/created from reactants.
c. Arrow – yields, produces, forms, creates, makes, becomes, turns into, etc.
i. Forward Reaction – reads from left to right. 
ii. Reverse Reaction – reads from right to left. 
d. Symbols (page 266)
i. Catalyst - chemical that speeds up the reaction without being used
up – neither a reactant nor a product – written above the arrow
Example:
MnO2
H2O2(aq)

H2O(l) + O2(g)
ii. Heat supplied to reaction. Greek letter Delta Δ written above
arrow.
iii. Pressure - written above the arrow (units = kPa, mmHg, atm)
iv. States of Matter
(s) solid
(l) liquid
(g) gas
(aq) aqueous – dissolved in water
Balancing Equations
a. Skeletal Equations - does not indicate the relative amounts of the
reactants and products – not balanced - uses element symbols and
chemical formulas rather than words. Just a list of ingredients.
Ex:
1. solid iron + oxygen gas  iron (III) oxide (rust)
Fe(s) + O2(g)  Fe2O3(s)
2. hydrogen peroxide  water + oxygen gas
H2O2(l)  H2O(l) + O2(g)
3. methane gas + oxygen gas  carbon dioxide gas + water vapor
CH4(g) + O2(g)  CO2(g) + H2O(l)
b. Law of Conservation of Mass – atoms are neither created nor destroyed,
they are just rearranged during a chemical reaction.
i. Balanced Chemical Equations - each side of the equation has the
same number of atoms (and, therefore, mass) of each element.
Example:
Burning solid carbon in the presence of oxygen gas produces
carbon dioxide gas.
C(s) + O2(g)  CO2(g)
c. Coefficients - small whole number that is written in front of a chemical
formula in a balanced chemical equation.
i. Coefficient applies to whole formula.
Box formula so you do not change it in any way.
ii. Can NEVER be a fraction of a whole number (no decimals).
iii. NEVER put a coefficient in the middle of a formula, only in front.
iv. NEVER change the chemical formulas subscripts.
d. Steps to balancing a skeletal equation (page 270-271)
i. Have the correct skeletal equation.
Check that your reactants are on the left.
Check that your products are on the right.
Check that all formulas are written correctly - net charges of zero.
Example:
Hydrogen gas and oxygen gas react to create water vapor.
H2(g) + O2(g)  H2O(g)
ii. List the elements in the reaction.
List the elements on reactant side under the reactants
Product side under the products in same order as reactants
List polyatomic ions as a whole group if they appear on both sides
Draw a vertical line from the arrow down, between reactants and
products, to keep them separate.
Example:
H2(g) + O2(g)  H2O(g)
H
O
H
O
iii. Count the atoms of each element in the reaction.
List the quantity of the reactants on the reactant side
Products on the product side in the same order as reactants
Remember to multiply through when you have parenthesis
Example:
H2(g) + O2(g)  H2O(g)
H
O
2
2
H
O
2
1
iv. Start to balance an element that appears in only one compound on
each side of the equation by placing a whole number coefficient in
front of a compound to increase the number of atoms of the element.
NEVER change the subscripts in the formulas.
Box the formula so you do not change it in any way.
If there is no coefficient written, then it is 1.
Re-Add atoms after placing a coefficient.
Remember to multiply through when adding atoms.
Example:
H2(g) + O2(g)  2H2O(g)
H
O
2
2
H
O
2 is now a 4 on product side
1 is now a 2 on product side
v. Repeat the last step for another element that appears only once on
each side of the equation. After each coefficient is added, keep
checking, by re-adding, to see if the atoms are balanced for each
element on both sides of the equation.
Example:
2H2(g) + O2(g)
H
O
 2H2O(g)
2 is now a 4 on reactant side
2 is still a 2 on reactant side
H
O
4
2
vi. Make sure all coefficients are in lowest whole number ratio.
Reduce when necessary!
Example:
If created… 4H2(g) + 2O2(g)  4H2O(l)
4:2:4
Then reduce to… 2H2(g) + O2(g)  2H2O(l)
2:1:2
NOTE: Avoid these two common MISTAKES….
1) Be sure to start off with the correct chemical formulas or nothing else
will be correct.
2) Never change the subscripts once you know that the formula is correct.
Balance equations by placing coefficients in front of the compound
formula, NOT by changing the subscripts in the formula!!!!
Chapter 9 Stoichiometry (Tuesday 1/11/11 P1,3,5 & Wednesday 1/12/11 P2,4,6)
IV.
Mole to Mole Calculation – Use mole ratios to do calculations requiring a
conversion from one type of compound to another.
a. Start with balanced chemical reaction. Recipe.
b. Use the coefficients as mole quantities to convert from one compound to
another by creating a mole ratio to use as the conversion factor.
c. Given: 2Al2O3(l)  4Al(s) + 3O2(g)
Question: The decomposition of five moles of Aluminum Oxide would
create how many moles of Oxygen Gas?
Note: From the balanced chemical equation, you already know the mole
ratio is 2mol aluminum oxide is needed to create 3mol of oxygen gas.
2mol Al2O3(l) = 3mol O2(g) is your conversion factor
So use…
V.
2mol Al2O3(l)
3mol O2(g)
Or
5mol Al2O3(l)
1
X
3mol O2(g)
2mol Al2O3(l)
Given over 1
X Conversion Factor
Is the Mole Ratio
From the Balanced
Chemical Reaction
3mol O2(g)
2mol Al2O3(l)
= 7.5mol O2(g)
= Answer
Stoichiometry – Now you can change from any units of one type of
compound into any units of another type of compound as long as they are
relate by a balanced chemical equation.
a. 1Mole = Formula Mass on the Periodic Table in grams
b. 1Mole = 6.02 x 1023 particles (molecules, formula units, atoms, ions)
c. 1Mole = 22.4L of a gas at STP (standard temperature and pressure)
d. Mole Ratio of Coefficients from the balanced chemical reaction.
e. Given: 2Al2O3(l)  4Al(s) + 3O2(g)
Question 1 (We will do together.): The decomposition of 325grams of
Aluminum Oxide would produce how many grams of Aluminum?
Question 2 (You try.): The decomposition of 3.4 x 1025 formula units of
Aluminum Oxide would liberate how many grams of Oxygen Gas?
Chapter 9 Percent Yield (Thursday 1/13/11 P1,3,5 & Friday 1/14/11 P2,4,6)
VI.
Percent Yield - the ratio of the actual yield to the theoretical yield multiplied
by 100 to express it as a percentage
a. Theoretical Yield –a math equation is used to calculate the product
b. Actual Yield – an actual lab experiment is used to create the product.
c. Where is the yield lost?
i. reaction may not have gone to completion
ii. impure reactants or “other” products
iii. filter or transfer loss
d. Formula:
Percent Yield =
Actual Yield
x 100
Theoretical Yield
Example: What is the percent yield of CaCO3 if 28 grams of CaCO3
produces 13.1 grams of CaO?
Second Semester starts here!!! (Wednesday 2/2/11 P1,3,5 & Thursday 2/3/11 P2,4,6)
Chapter 8 Types of Chemical Reactions
VII. Types of Chemical Reactions - Doing reactions in a lab can be dangerous,
time consuming and/or expensive. Recognizing patterns allows scientists to
predict the products of many reactions.
a. Synthesis – also called combination reactions. Two or more reactants
form a single product. Reactants are usually two elements or two small
compounds that combine. Product must be a single compound. Try to
create acids or polyatomic ions in product.
E+EC
or
small C + small C  larger C
i. Elements plus Oxides 2Mg(s) + O2(g)  2MgO(s)
2H2(g) + O2(g)  2H2O(l) Hindenburg
ii. Metals with Halogens
2Na(s) + Cl2(g)  2NaCl(s)
iii. Active Metals with Oxides CaO(s) + H2O(l)  Ca(OH)2(s)
b. Decomposition – also called analysis reactions. A single reactant is
broken down into two or more products. Reactant is a single compound.
Products can be two or more elements or simpler compounds. Try to
create diatomic gases and water. These reactions usually require energy
(heat, light, electricity) to happen.
CE+E
or
large C  smaller C + smaller C
i. Binary Compounds
Δ
2HgO

2Hg
+ O2(g)
Δ
ii. Metal Carbonates
CaCO3(s)

CaO(s) + CO2(g)
Δ
iii. Metal Hydroxides
Ca(OH)2(s) 
CaO(s) + H2O (g)
Δ
iv. Metal Chlorates
2KClO3(s) 
2KCl(s) + 3O2(g)
MnO2 (catalyst)
Δ
v. Acids
H2SO4(aq) 
SO3(g)
+ H2O(l)
c. Single-Displacement - atoms of one element replace the atoms of a
second element in a compound. Replacement. (Dancing couple and a
single person cuts in.) Reactants are a compound and an element. Products
are a different compound and element. Cation (+) replaces the other cation
(+) and anion (-) replaces the other anion (-).
A + BC  B + AC
or
A + BC  C + BA
(+)
(+)(-)
(+)
(+)(-)
(-)
(+)(-)
(-)
(+)(-)
i. Displacement of Metal by Metal - Activity Series of Metals (pg.
286) - More Reactive Metals Replace the Less Reactive Metals
Mg(s) + Zn(NO3)2(aq)  Mg(NO3)2(aq) + Zn(s)
Mg is more reactive than Zn.
Mg(s) + LiNO3(aq)  No Reaction
Mg less reactive than Li.
ii. Displacement of Hydrogen in Water by Metal
Na(s) + H2O(l)  NaOH(aq) + H2(g) (not balanced)
Visualize H2O as H+ and OHiii. Displacement of Hydrogen in Acid by Metal
Mg(s) + 2HCl(aq)  H2(g) + MgCl2(aq)
iv. Displacement of Halogens - Diatomic Elements in Group 7A - Top
elements more reactive than bottom elements.
Cl2(g) + 2KBr(aq)  2KCl(aq) + Br2(g)
Cl is more reactive than Br.
d. Double-Replacement - involves the exchange of positive ions between
two compounds. (Like square dancing – when both couples switch
partners) Usually happens between two ionic compounds in water
(aqueous). Reactants are two ionic compounds. Products are two different
ionic compounds.
AB + CD 
AD + CB
(+)(-)
(+)(-)
(+)(-)
(+)(-)
i. Formation of a Precipitate – solid – cloudy white
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
ii. Formation of a Gas – bubbles
FeS(s) + 2HCl(aq)  FeCl2(s) + H2S(g)
iii. Formation of Water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
e. Combustion - combusts, explodes, burns, etc. Reactants must include
oxygen gas and usually includes a compound with carbon (fuels). If
remove the oxygen and smother the reaction. Products usually are carbon
dioxide gas and water vapor. Reaction usually also produces energy in the
form of heat and/or light (noise, electricity).
CxHy + O2(g)  CO2(g) + H2O(g) or CxHyOz + O2(g)  CO2(g) + H2O(g)
i. Production of Water
2H2(g) + O2(g)  2H2O(l) The Hindenburg
ii. Production of Water and Carbon Dioxide
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
VIII.
Predicting Products
a. Combination – one product – create a polyatomic or acid
Compound + Compound  bigger Compound
Element + Element  Compound
b. Decomposition – one reactant - create gases and liquids
Compound  Element + Element
Compound  smaller Compound + sm. Compound
c. Single Replacement – Dancing couple + single guy
The more reactive element replaces the less reactive element.
E + C  Different E + Different C
Cation (+) replaces Cation (+) OR Anion (-) replaces Anion (-)
d. Double Replacement – Square dancing
C(aq) + C(aq)  Different C(s or g) + Different C(aq)
Cation (+) replaces Cation (+) AND Anion (-) replaces Anion (-)
e. Combustion – O2(g) must be reactant, CO2(g) & H2O(g) are usually products
CxHy + O2(g)  CO2(g) + H2O(g)
CxHyOz + O2(g)  CO2(g) + H2O(g)
f. Reminders:
Look for polyatomic ions and acids.
Look for water and diatomic gases.
H20 breaks apart into H (1+) and OH (1-).
Make sure all diatomic elements have the subscript 2.
Make sure all ionic formulas have a net charge = zero.
Make sure all equations are correctly balanced.
Chapter 9 Limiting Reactants (Tuesday 2/8/11 P1,3,5 and Wednesday 2/9/11 P2,4,6)
IX.
Limiting Reactant - the reactant (ingredient) that runs out first – determines
the amount of possible product.
a. Excess Reagent - the reactant (ingredient) that is left over after all of the
other reactant is used up in the creation of the product. You must compare
the given quantity to the required quantity of each reactant.
b. COOKIES: The cookie recipe says that I need 1 cup of sugar and 2 cups
of flour to make 1 batch of cookies. What the ratio of sugar to flour?
If I had 5 cups of sugar and 15 cups of flour in the cupboard...
How much flour to use all 5 cups of sugar? What is the ratio?
How much sugar to use all 15 cups of flour? What is the ratio?
Which is possible, to use all the sugar or all the flour?
How many batches of cookies could I make?
Which ingredient is limiting?
Which ingredient will be left over? How much will be left over?
c. Steps:
i. Find how much product can be made by each given amount
reactant
ii. Determine the limiting reagent
iii. The limiting reagent controls the possible product quantity
iv. Determine the quantity of left over excess reagent
Example: What will occur when 6.7 mol of solid sodium reacts with 3.2 mol
of chlorine gas? Which is the limiting reagent? How many moles of sodium
chloride are produced? How much of the excess reagent remains?
Example: What is the maximum number of grams of Cu2S - Copper (I)
Sulfide - that can be formed when 80.0 grams of copper react with 25.0 grams
of sulfur?