Download Chapter 4:Chemical Quantities and Aqueous Reactions:

Chapter 4
Chemistry 1A
Page |1
Chapter 4: Chemical Quantities and Aqueous Reactions:
Homework: Read Chapters 4. Work out sample/practice exercises
Bonus Chapter 4 Problems: 25, 31, 37, 43, 49, 53, 57, 59, 63, 67, 69, 73, 77,
79, 83, 87, 89, 93, 95, 101
Check MasteringChemistry.com assignment and complete before due date
Reaction Stoichiometry: Chemical arithmetic.
Stoichiometry is the study of
numerical relationships between chemical quantities in a chemical reaction. Using ratios
found in balanced chemical reactions to solve quantitative problems.
Convert to moles, mole to mole ratio, convert to desired result.
Practice:
1.
Balance the equation below and calculate the mass in grams of water vapor
produced when 56.0 g of pure hydrogen gas burns in an unlimited supply of oxygen.
H2 (g) +
O2 (g) 
H2O (g)
2.
How many mg of solid iron (III) chloride (FeCl3) result when 15.5 milligrams of
solid iron (Fe) is reacted with an excess of chlorine gas (Cl 2). Start by balancing the
equation.
Chapter 4
Chemistry 1A
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Yields:
Actual or Experimental Yield = that produced in a lab environment.
Theoretical Yield = the highest (maximum) possible yield assuming 100%
perfection using up all of the limiting reactant.
Percent Yield =
x 100%
Limiting reactants:
The limiting reactant is completely consumed producing the least amount of
product. It runs out first and limits the amount of product.
It produces the lowest (maximum) theoretical yield. Limiting reactant
problems occur when two values for reactants are given.
Excess reactants:
Excess reactants are not completely consumed in the reaction.
Practice:
3.
a)
Balance the following unbalanced equation:
BaO (s) +
4.
HCl (aq) 
BaCl2 (aq) +
H2O (l)
b)
There are 2 moles of BaO and 2 moles of HCl, which is the limiting reactant?
a)
Balance the following unbalanced equation:
Ca (s) +
b)
c)
d)
H2O (l) 
Ca(OH)2 (aq) + H2 (g)
How many grams of H2O would be required to react with 8.00 g of calcium?
Solve for the theoretical mass of hydrogen gas produced.
When the reaction described above was carried out in the lab, 0.346 grams of
hydrogen gas was obtained. Calculate the percent yield for this reaction.
Chapter 4
5.
a)
c)
d)
Page |3
Balance the following unbalanced equation:
P4 (s)
b)
Chemistry 1A
+
F2 (g) 
PF5 (g)
Calculate the theoretical mass in grams of phosphorus pentafluoride
produced from
12.6 grams F2 and 9.00 g phosphorus.
What is the limiting reactant from the reaction above?
What is the % yield of the reaction above if 15.8g of PF 5 were experimentally
produced in a lab starting with the given amounts above?
Solution Concentrations:
Homogenous mixtures are called solutions. The major component that keeps its
same state is called the solvent. The minor component, the one that may have a
change in state, is the solute.
Dilute solutions have a small amount of solute, concentrated solutions have a large
amount of solute compared to the solvent.
Molarity = (moles of solute)/(liter of solution)
Molarity can be used as a conversion factor to solve for moles in
stoichiometry problems
Solution Dilution: M1V1 = M2V2
When diluting solutions it is best to use a volumetric flask and safest to
start with some solvent, then add the concentrated amount, followed by
more solvent to reach the desired total volume.
Note: Volumes are not always additive: for example 50 ml of water
plus 50 ml of ethanol only produce 95 ml volume total due to some
differences in density and intermolecular attractions
Other concentrations for solutions include: mass percent, volume percent,
mass/volume percent, ppm, ppb, normality, molality, mole fraction.
Chapter 4
Chemistry 1A
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Practice:
6.
Calculate the Molarity of a solution made from 30.0 grams of sucrose (C12H22O11)
added to enough water to make 300 ml total solution.
7.
Describe quantitatively and qualitatively how one should safely prepare 500 ml of
1.0 M H2SO4 (aq) solution from 18.0 M H2SO4 stock solution.
8.
How many grams of hydrogen gas will be theoretically formed by the reaction of
10.0 ml of 6.00 M HCl with excess zinc?
Zn (s)
+
HCl (aq)

ZnCl2 (aq) +
H2 (g) unbalanced
Chapter 4
Chemistry 1A
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Titrations:
In a titration, a substance in a solution of known concentration is reacted with
another of unknown concentration. A burette is used to add a known volume to a
flask with an indicator that will change (endpoint) when the reactants reach their
stoichiometric ratio (equivalence point).
Practice:
9.
Solve for the molarity concentration in a 10.0 ml portion of acetic acid solution
(HC2H3O2) that required 39.6 ml of 0.202 M NaOH solution to reach the
equivalence point in a titration.
Chapter 4
Chemistry 1A
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Gas Stoichiometry:
Molar volume at STP for any ideal gas is 22.4 L/mol
STP = standard temperature (0°C) and pressure (1 atmosphere)
Practice:
10. Pure iron may be prepared from iron(III) oxide and carbon monoxide:
Fe2O3 (s) +
CO (g) 
Fe (s) +
CO2 (g)
a) Balance the unbalanced equation above.
b) Calculate the maximum theoretical mass in grams of Fe produced from 40.0 L
of carbon monoxide at STP from the reaction above.
Dissolving a Solute
When mixing a solute with the solvent, attractions between the like substances must
break while attractions between solute-solvent are formed. If the new solute-solvent
attractions are strong enough, the solute will dissolve.
Water is a polar molecule with
uneven distribution of
electrons: (more on the oxygen
-, less on the hydrogen +)
When ionic compounds
dissolve in water, each ion is
attracted to the polar water and
surrounded by a solvent cage of
water. The result is a solution
with free moving charged
particles that may conduct
electricity.
Materials that dissolve in water to form a solution that will conduct electricity are
called electrolytes.
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Electrolytes in Aqueous Solution:
Strong Electrolytes will largely dissociate into its ions in an aqueous solution and
are written as separated ions in the ionic reactions.
Examples: Strong acids, Strong Bases, Soluble salts.
Strong Acids: HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3
Strong Bases: soluble hydroxides from Group 1A (not including H)
and Group 11A, not including the top two) LiOH, NaOH, KOH,
RbOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
Soluble Salts: Ionic compounds that contain the cations from Group
1A; Li+, Na+, K+, Rb+, Cs+, or ammonium ion, NH4+. A compound is
probably soluble if it has the anion; Cl-1, Br-1, I-1 except with Ag+,
Hg2+2, or Pb+2, and most compounds that include NO3-1, ClO4-1,
C2H3O2-1, Soluble with most SO4-2 except Ba+2, Hg2+2, or Pb+2.
Weak Electrolytes will partially dissociate into its ions in an aqueous solution, but
are written as compounds in an ionic equation. Weak electrolytes are the weak
acids and weak bases such as HC2H3O2 or NH3
Nonelectrolytes will stay in their formula in an aqueous solution. Nonelectrolytes
are the insoluble salts, gases, liquids, and other solids.
Writing Molecular, Ionic, and Net ionic equations:
Example: For the double displacement reaction of lead (II) nitrate reacting
with aluminum chloride.
Whole equation: Unbalanced
Pb(NO3)2 (aq) + AlCl3 (aq)  PbCl2 (s) + Al(NO3)3 (aq)
Total ionic equation: Unbalanced
Pb+2(aq) + NO3-1(aq) + Al+3(aq) + Cl-1(aq)  PbCl2 (s) + Al+3(aq) + NO3-1(aq)
Identify the spectator ions and leave them out when writing the net ionic eq.
Net ionic equation: Unbalanced
Pb+2 (aq) + Cl-1 (aq)  PbCl2 (s)
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Chemical Reactions:
Evidence of chemical reactions:
Examples…formation of a solid, or formation of gas, light given off, change in
temperature, etc.
Driving forces of chemical reactions:
1)
Transfer of electrons occurs in redox (oxidation-reduction) reactions
2)
Formation of solid
3)
Formation of gas
4)
Formation of water or a weak electrolyte (weak acid or weak base)
5)
Other driving forces not listed may occur
Classifying Chemical Reactions:
Types of Chemical Reactions:
(D-D) double displacement: AB + CD  AD + CB
(P) precipitation: a solid is formed
(N) neutralization or acid-base: water is formed
(G) gas evolution: gas is formed
(Syn) synthesis or combination: A + B  AB
(D) decomposition: AB  A + B
(R) redox or oxidation-reduction: electrons are transferred
(C) combustion: add O2 gas, possible products include: CO2, H2O, NO2, SO2.
(SR) single replacement: A + BC  AC + B or BA + C
The more active metal will have a partner, less active is alone.
Activity series: Au<Hg<Ag<Cu<H2<Pb<Sn<Ni<Cd<Fe<Cr< Zn<Mn<Al< Mg< Na<Ca< K<Li
Chapter 4
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Classifying and Predicting Products and Chemical Reactions:
The following examples will help you classify the type of reaction and predict the
expected products. Refer to the solubility rules to determine if solid forms.
(D-D) double displacement: AB + CD  AD + CB
A double displacement reaction starts with two ionic compounds in which the ions
exchange to produce new balanced ionic compounds. Always write the cation
before the anion and verify that the net charge of a compound is zero. Ionic
compounds may be acid, base, or salt. Oxidation charges do not alter in double
displacement reactions so this type will never be a redox reaction.
(NR) When all ionic products are strong electrolytes, there is NO REACTION.
FeBr3 (aq) + 3 LiNO3 (aq)
 3LiBr (aq) + Fe(NO3)3 (aq)
NR
Driving forces for double displacement reactions include…
(P) Precipitation, formation of solid
(N) Neutralization or Acid-Base Reaction, formation of water
(G) gas evolution, formation of a gas
(WE) formation of a weak electrolyte, a soluble weak acid or weak base
(P)
Formation of a new solid in a double displacement reaction when there is
no solid on the reactant side is called PRECIPITATION.
Ni(NO3)2 (aq) + Na2S (aq)
(N)

NiS (s) +
2 NaNO3 (aq)
DD,P
Formation of water in a double displacement reaction is called
NEUTRALIZATION (also known as Acid-Base reaction)
3 KOH (aq) + H3PO4(aq)

K3PO4 (aq) + 3 H2O (l)
DD,N
(WE) Formation of a WEAK ELECTROLYTE, generally a weak acid, produced
from a double displacement reaction
Al(C2H3O2)3 (aq) + 3 HCl (aq)  AlCl3 (aq) + 3 HC2H3O2 (aq)
DD,WE
(G) Gas Evolution:
Several common products are in the gas phase. You should recognize the following
as gas products.
CO2, SO2, NH3, H2S, H2, N2, O2, F2, Cl2, NO2
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Special cases of (G) Gas Evolution in (DD) Double Displacement:
There are three common products that further decompose leading to three products
in place of just the two products in a double displacement reaction. They are as
follows.
NH4OH (aq)
 NH3 (g) + H2O (l)
H2SO3 (aq)
 SO2 (g) + H2O (l)
H2CO3 (aq)
 CO2 (g) + H2O (l)
The double displacement reactions below show examples of the decomposition of
one of the ionic products. The driving forces include both formation of water and a
gas, because water is formed these can be labeled DD, N, G reactions. The last one
is DD, N, G and P.
K2CO3 (aq) + 2HCl (aq) 
H2CO3 (aq) + 2 KCl (aq)
K2CO3 (aq) + 2HCl (aq)  CO2 (g) + H2O (l) + 2 KCl (aq)
Na2SO3 (aq) + 2HCl (aq)
H2SO3 (aq) + 2 NaCl (aq)
Na2SO3 (aq) + 2HCl (aq) SO2 (g) + H2O (l) + 2 NaCl (aq)
(NH4)2SO4 (aq) + Ba(OH)2 (aq)
2 NH4OH (aq) + BaSO4 (s)
(NH4)2SO4 (aq) + Ba(OH)2 (aq) 2 NH3 (g) + 2 H2O (l) + BaSO4 (s)
(Syn) Synthesis or combination: A + B  AB
A synthesis reaction starts with two or more substances and creates a single
compound product. The reverse of a synthesis reaction is Decomposition.
Remember which elements are diatomic elements: H2, N2, O2, F2, Cl2, Br2, I2. Know
the elements in their pure form: gas, solid, and liquid.
a) two elements (metal plus nonmetal)  ionic compound.
K (s) + Cl2 (g)  KCl (s)
b) metal oxide and water  base (metal hydroxide)
BaO (g) + H2O (l)  Ba(OH)2 (aq)
c) nonmetal oxide and water  acid
CO2 (g) + H2O (l)  H2CO3 (aq)
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(D) Decomposition: AB  A + B
A decomposition reaction starts with one reactant and breaks up into two or more
substances. The reverse of a decomposition reaction is synthesis. Remember the
diatomic elements: H2, N2, O2, F2, Cl2, Br2, I2.
a) ionic compound  two elements (metal plus nonmetal)
2 Fe2O3 (s)  4 Fe (s) + 3 O2 (g)
b) metal carbonate  metal oxide and carbon dioxide
CaCO3 (s)  CaO (s) + CO2 (g)
c) metal bicarbonate  metal carbonate + water + carbon dioxide
2 Al(HCO3)3 (s)  Al2(CO3)3 (s) + 3 H2O (l) + 3 CO2 (g)
d) metal connected to halogen with oxygen  metal halide + oxygen
Ca(BrO3)2 (s) CaBr2 (s) + 3 O2 (g)
e) hydrated salt  anhydrous salt plus water
CuSO4.5H2O (s)  CuSO4 (s) + 5 H2O (l)
(R) Redox or oxidation-reduction:
Redox reactions have a transfer of electrons. An easy what to tell that a redox
reaction takes place is to check for element charges and see if they change from
reactant to product. Some synthesis and decomposition reactions are redox such as
the synthesis (a), and decomposition (a), (d) reactions. All single replacement and
combustion reactions are redox reactions.
(C) Combustion: elements burn with O2 gas to produce oxides,
Common products include: CO2, H2O, NO2, SO2.
NH3 (g) + O2 (g) 
C4H10 (s) + O2 (g) 
C17H24S (s) + O2 (g) 
C12H22O11 (s) + O2 (g) 
(SR) Single Replacement: A + BC  AC + B or BA + C
The more active metal will have a partner, less active is alone.
Activity series: Au<Hg<Ag<Cu<H2<Pb<Sn<Ni<Cd<Fe<Cr< Zn<Mn<Al< Mg< Na<Ca< K<Li
Do NOT memorize the activity series. The necessary information will be
given to you as needed. Most of the transition metals commonly have a +2
charge, except Ag+1, Cr+3, and Au+3
Chapter 4
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Predict the products if a reaction occurs or write NR for no reaction…
Zn(NO3)2 (aq) + Cu (s) 
Cu(NO3)2 (aq) + Ag (s) 
AgNO3 (aq) + Cu (s) 
The nonmetal activity levels increase as you go from the bottom to the top of the periodic
table:
Given that F>Cl>Br>I. Predict the products or write NR…
NaBr (s) + Cl2 (g) 
NaBr (s) + I2 (s) 
Redox Reactions:
Redox reactions exchange electrons. When balancing a reaction …
(1) electrons lost = electrons gained
(2) atoms on either side of the reaction must balance
(3) total charge on either side of the reaction must balance.
Oxidation
increases oxidation number
lose electrons
2 Cl-1  Cl2 + 2 eCl-1 is oxidized
Cl-1 is a reducing agent
Reduction
reduces oxidation number
gain electrons
K+ + 1 e-  K
K+ is reduced
K+ is an oxidizing agent
Rules for Assigning Oxidation Numbers
1)
elements
zero
2)
monatomic ions
charge on ion
3)
oxygen
mostly (-2)
Exceptions:
peroxide, O2-2, (-1) and
superoxides, O2-1, (-1/2)
4)
hydrogen
mostly (+1)
exceptions:
hydrides (-1)
5)
halogens
mostly (-1)
Exceptions: (+1, +3, +5, +7)
in a compound with lighter halogens or oxygen
6)
Compounds
sum of oxidation numbers equal zero
7)
polyatomic ions
sum of oxidation numbers equal charge of ion
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Balancing Complicated Redox Rxns using the Half Rxn Method (Ch 18):
The half reaction method of balancing for the electrochemistry reactions splits apart
the oxidation half reaction from the reduction half reaction and then adds the two
together for an overall reaction.
1)
Assign oxidation numbers to each atom.
2)
Determine what is being oxidized and what is reduced.
3)
Split the skeleton equation into two half reactions label appropriately as oxidation
or reduction reactions.
4)
Complete and balance each half reaction (This method does not need to use the
oxidation numbers)
a)
balance atoms that change their oxidation numbers
b)
balance O by adding H2O
c)
balance H by adding H+
d)
balance charges by including the electrons as products in the oxidation
reaction and electrons as reactants in the reduction reaction
e)
If the solution is basic: add equal OH-1 ions on both sides of the equation to
counter all H+ ions, every H+ and OH-1 become H2O essentially removing all
H+ ions from the reaction.
5)
Combine the two half reactions to obtain overall equation
a)
Multiply each half reaction by the factor needed to
have equivalent
electrons lost and gained.
b)
Simplify and reduce to the lowest whole number ratio
6)
Check balancing
a)
Do atoms balance?
b)
Are the sum of charges of reactants equal to those of products?
c)
Are the same number on electrons lost and gained?
Redox: Balance the following.
a) Al (s) + SO4-2 (aq)  Al+3 (aq) + SO2 (g)
(acidic conditions)
b) A breathalyzer detects ethanol in suspected drunk drivers using the reaction below
Cr2O7-2 (aq) + C2H5OH (l)  Cr+3 (aq) + C2H4O (l) (acidic conditions)
Chapter 4
c) MnO4
d) MnO4
e) Br 2
-1
-1
Chemistry 1A
+2

MnO2
-1

Mn
+
Mn
+
IO3

BrO 3
-1
+
+2
P a g e | 14
(acidic conditions)
+
Br
IO4
-1
-1
(basic conditions)
(basic conditions)
Redox Applications: Combustion, Bleaching, Batteries, Metallurgy, Corrosion,
Respiration, Titration
Redox Titration: Electrons transfer. Titrate with burette as liquid is slowly added
to a flask. Color changes to indicate endpoint just like acid/base titrations.
Example :
The indicator, MnO4-1 has an intense dark purple color. For the example below,
MnO4-1 is a dark purple reactant in the burette. It will colorlessly react with oxalic
acid in the flask until the oxalic acid runs out. At the endpoint, excess MnO4-1 turns
the flask solution purple.
Calcium levels in blood may be determined by adding oxalate ion to precipitate the
calcium oxalate, CaC2O4…
Ca+2(in blood) + C2O4-2  CaC2O4 (s) (solid is separated out of the blood)
followed by dissolving the precipitate in acid…
CaC2O4 (s) + 2H+1 (aq)  H2C2O4 (aq) + Ca+2
and titrating the resulting oxalic acid with KMnO4 (K+ is a spectator ion which is
left out of the reaction.)
5H2C2O4 (aq) + 2MnO4-1(aq) + 6H+1(aq)  10CO2(g) + 2Mn+2(aq) +8H2O(l)
How many milligrams of Ca+2 are present in 10.0 ml of blood if 21.08 ml of
0.00988M KMnO4 solution is needed?
Chapter 4
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Memorize the Soluble Solubility Rules:
Compounds containing these ions are Exceptions: (when combined with
generally soluble
ions on left, compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3-, C2H3O2-
none
Cl-, Br-, I-
Ag+, Hg2+2, Pb+2
SO4-2
Ag+, Ca+2, Sr+2, Ba+2, Pb+2
Be able to use the Insoluble Solubility Rules:
Compounds containing these ions are Exceptions: (when combined with ions
on left, compound is soluble or slightly
generally insoluble
soluble
OH-
Li+, Na+, K+, NH4+
Ca+2, Sr+2, Ba+2
S-2
Li+, Na+, K+, NH4+, Ca+2, Sr+2, Ba+2
CO3-2, PO4-3
Li+, Na+, K+, NH4+
Practice:
11. List elements that are diatomic.
12. Name several driving forces for a chemical reaction.
13. Name several signs of evidence that would indicate a chemical reaction happened.
14. What information can one obtain from a balanced chemical equation?
15. a)
Complete and balance the double-displacement whole equation.
Include the physical states.
Na2SO4 (aq) +
AgNO3 (aq) 
b)
Write the balanced total ionic equation for the above reaction. Include
charges on ions and physical states. Circle the spectator ions.
c)
Write the balanced net ionic equation for this reaction.
d)
What is/are the driving force(s) for this reaction?
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16. Crude gunpowders often contain a mixture of solid potassium nitrate and charcoal
(solid carbon). When such a mixture is heated until a reaction occurs, a solid
residue of potassium carbonate is produced. The explosive force of the
gunpowder comes from the fact that two gases are also produced, carbon monoxide
and nitrogen, which increase in volume with great force and speed. Write the
balanced chemical equation for the reaction.
17. Balance and Classify in as many ways possible the following reactions.
Possible classifications:
(R) redox or oxidation-reduction
(D) decomposition
(N) neutralization
(C) combustion
(G) gas evolution
(D-D) double displacement
(SR) single replacement
(Syn)synthesis or combination
(P) precipitation
a)
NaI (s) + F2 (g) 
b)
NH3 (g) +
c)
N2 (g) +
d)
C4H10 (l) + O2 (g)
e)
As(OH)3 (aq) + H2SO4(aq)  As2(SO4)3(aq)
f)
aqueous silver nitrate and solid copper reacts to form aqueous copper (II)
nitrate and solid silver.
g)
solid magnesium chlorate heated produces oxygen gas and solid magnesium
chloride
NaF (s) + I2 (s)
O2 (g) 
O2 (g) 
NO2 (g)
+
H2O (l)
N2O5 (g)

CO2 (g)
+
H2O (g)
+ H2O (l)
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P a g e | 17
h)
A chemical process that uses electricity is called an electrolysis reaction.
When an electrical current passes through a concentrated aqueous sodium
chloride solution, several useful products can be isolated: aqueous sodium
hydroxide, chlorine gas, and hydrogen gas. Write the balanced chemical
equation for this process.
i)
Fe (s)
j)
H2SO4(aq) + Na2CO3(aq)  Na2SO4(aq) +
k)
Al2(CO3)3 (s)
l)
HC2H3O2 (aq) + Mg (s) 
m)
Ni(NO3)2 (aq) + K2S (aq)

NiS (s) +
KNO3 (aq)
n)
KOH (aq) + (NH4)2SO4(aq) 
NH3 (g) +
K2SO4(aq)
+
Br2 (l) 

FeBr3 (s)
H2O(l) + CO2(g)
Al2O3 (s) + CO2 (g)
H2 (g) + Mg(C2H3O2)2(aq)
+
H2O (l)
18. For the following reactions below: Complete and balance the reactions. Indicate
what the driving force(s) of the reaction will be. If there are no driving forces write
NR for no reaction.
a & b are double displacement; c is synthesis.
DRIVING FORCE(S)
a)
PRODUCTS
HBr (aq) + Li2CO3 (aq) 
b)
Pb(NO3)2 (aq) + (NH4)2S (aq) 
c)
Cl2 (g) +
Al (s)

+
+
+
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19. Complete and Balance the following chemical reactions.
Include the state that the substances are in, (s), (l), (g), or (aq). If no reaction in a
double displacement then write No reaction, but balance it anyway.
a)
Combustion reaction:
C7H16S2 (s) +
b)
O2 (g)

Single replacement reaction: (Zn > Cu)
Solid zinc is added to aqueous copper (II) chloride
c)
Decomposition reaction:
Fe(HCO3)3 (s) 
d)
Double Displacement reaction:
aqueous sulfuric acid is added to aqueous ammonium carbonate
e)
Synthesis reaction:
CO2 (g) + H2O (l) 
f)
Double displacement reaction:
KNO3 (aq) +
g)
(NH4)3PO4 (aq) 
Double displacement reaction:
aqueous acetic acid + aqueous barium hydroxide 
h)
Decomposition:
CaCO3 (s) 
j)
Neutralization:
HCl (aq)
+ K2SO3 (aq) 