UNIT 7 CP CHEM NOTES - CHEMICAL EQUATIONS
I. Chemical Reactions and their equations.
1. The materials which you can start with are written first, and are called reactants. If
there is more than one reactant, a plus (+) sign separates each individual reactant.
Example: Sodium reacts with chlorine to form sodium chloride (NaCl). An arrow is
written after the reactants.This is how the reactants would be written...
Na + Cl2 ------->
2. The product(s) are those substances which are formed or produced in the reactions.
They are always different substances than what you start with.
Na
+
Cl2
-------->
NaCl
3. If atoms are unbonded a plus sign separates them. In the above example, the plus sign
tells you that there are two reactants, and they are not bonded together.
4. To indicate the number of atoms (unbonded) or the number of molecules, you use the
coefficient. In the final equation on the left side the 2 in front of the Na indicates that 2
atoms, unbonded, of sodium were required for this reaction. The 2 in front of the sodium
chloride indicates that 2 separate molecules of sodium chloride were formed. There is an
implied 1 for the coefficient of the chlorine...
2Na
+
Cl2
--------> 2NaCl
5. If the atoms are bonded, they are not separated by a plus sign. In the example, the
product, sodium chloride, indicates that the sodium and chlorine are bonded together into
a molecule.
6. To indicate the number of each atom bonded into a molecule, a subscript is needed.
In the above example, the chlorine consists of two atoms of Cl bonded. If it had been
written 2Cl, we would still have two atoms of chlorine, but they would not be bonded
together.
7. When balancing an equation, you must be sure to have the same number of atoms of
each element on the left of the arrow as you have on the right side of the arrow. Law of
conservation of atoms: Atoms may neither be created nor destroyed. If you look at
the equation written in the number 2 above, you would note that there is ONE Na indicated
on the left, and ONE Na indicated on the right. However there are two Cl atoms on the
left, but only one on the right. This form of the equations violates the above law. You
must now employ the process known as BALANCING EQUATIONS.
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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**** When balancing an equation you may never change the subscripts
(once the correct formula for the compounds are written). You may change
the coefficients to any number you wish in order to balance the
equation. When you think you are finished balancing, check to see if
all of the coefficients can be divided by the same number. If not, make the
correction.
Examples:
Na
+
Cl2
------->
NaCl
A. In order to balance this equation you may NOT:
1. Erase the subscript of the Cl. (Na + Cl ----> NaCl)
2. Add a subscript to the NaCl molecule to make Na Cl2
B. You should say to yourself, “I need to have two chlorine atoms on the right”.
You may put a coefficient 2 in front of the entire NaCl formula. You may not
put a 2 in between the Na and the Cl. The balanced equation reads...
Na
+
Cl2
------->
2NaCl
C. The 1 in front of the NaCl (or any coefficient in the front of a compound)
multiplies the number of all the atoms in the compound by that
number. This now means that you have one atom of sodium on the left
and TWO atoms of sodium on the right.
D. To correct this situation, which violates the law of conservation of atoms, go
back to the left side of the equation and put a 2 in front of the Na. This only
multiplies the Na by 2 since there is a plus sign separating the Na and the Cl.
The equation now reads...
2Na
+
Cl2
--------> 2NaCl
You now have 2 Na atoms on the left, and two atoms on the right side. You
also have 2 Cl atoms on the left, and two Cl atoms on the right. When
balancing equations it is not necessary to have the same number of each
element, but the number of atoms of each element on the left, must equal the
same number of atoms of that element on the right.
II. Practice balancing chemical equations...
A. The process of rusting:
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Fe
+
O2
------>
Fe2O3
B. Used as an oxygen source in explosives:
C.
KClO3
------->
O2
+
KCl
MgBr2 +
Cl2
-------->
MgCl2
+
-------->
CO2 +
H2O
------->
NaOH
+
O2
+
HCl
Br2
D. A basic fuel for high performance engines:
C2H6
+
O2
E. A solid oxidant for rocket fuels:
Na2O2
+
H2O
F.
KIO
------->
G.
NaCl
+
H2SO4
H.
Ag
+
HNO3
KI
+
------>
-------->
KIO3
Na2SO4
AgNO3
+
NO2 +
H2O
Helpful Hints
Balance elements last
Treat polyatomic groups as packages
For combustion reactions - use “CHO” to have sequence in balancing each element
Worksheet EE CP Chem
Name________________
Write and Balance simple Equations
I. Memorize the Diatomic Elements: H2 Br2 O2 N2 Cl2 I2 F2
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Mnemonic devices :
HBrONClIF
HNFOIClBr
HOClBrFIN
(havenofearoficecoldbeer)
(Huckleberry Fin)
II. Sometimes it is necessary to abbreviate the “phase” of the substance to the lower right of the substance.
Some abbreviations are: s = solid, l= liquid, g (
III Write equations using correct formulas, then Balance the equation for each of the reactions.
1. Hydrogen gas reacts with chlorine gas to yield Hydrogen chloride.
2. Carbon reacts with oxygen gas to form Carbon dioxide.
3. Lithium reacts with chlorine gas to form Lithium Chloride.
4. Calcium reacts with Bromine to yield Calcium Bromide.
5. Hydrogen gas reacts with oxygen gas to yield water.
6. Water decomposes to form Hydrogen gas and Oxygen gas.
7. Carbon Dioxide decomposes to form carbon monoxide and oxygen gas.
8. Hydrogen gas reacts with nitrogen gas to yield Ammonia gas (NH3)
9. Aluminum metal reacts with oxygen gas to form Aluminum oxide.
10. Nitrogen gas reacts with Iodine to yield Nitrogen triiodide
11. zinc + chlorine ---> zinc chloride
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Notes- Five General Categories of Chemical Reactions
I Composition - ( Also called Combination or Synthesis ) A complex substance is formed when two
or more smaller substances chemically combine.
General Forms:
Element
or
+
Element --------> Compound
Simple Nonionic Compound + Simple Nonionic Compound ----> Complex compound
Examples:
C
+
O2
----------->
CO2
Fe
+
S
----------->
FeS
2Mg
+
O2
----------->
2MgO
NH3
+
H2O
----------->
NH4OH
II Decomposition- A Single compound is broken into two or more smaller
substances, either elements or compounds
General Forms:
ONE Complex Compound
-------- > Simple compound +
or element
Simple compound
or element
Examples:
2H2O
----------->
O2
+
2H2
2KClO3
----------->
2KCl +
3O2
2HgO
----------->
Hg
+
O2
CaCO3
----------->
CaO
+
CO2
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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III Single Replacement- A Reaction between an element and an ionic compound, to form a new
element and a new ionic compound. If the element reacting forms positive ions , it replaces the positive
part of the ionic compound. If the element forms negative ions , it replaces the negative part of the
compound.
General Forms:
Element
+
Ionic compound
-------> new Ionic Compound
+ new element
Examples:
Fe
+
CuSO4 ----------->
Cl2
+
2KI
Zn
+
H2SO4 ----------->
Cu
+
2AgNO3
FeSO4 +
----------->
2KCl
ZnSO4 +
----------->
Cu
+
I2
+
2Ag
H2
Cu(NO3)2
IV Double Replacement- ( Also called Ionic) A Reaction between one ionic and a second ionic
compound, to form two new ionic compounds. The positive ions of the ionic compound switch
places. The negative ions ,switch places.
General Forms:
ionic Compound + Ionic compound -------> new (Ionic )Compound
+ new Ionic Compound
Examples:
BaCl2
H2SO4 ----------->
BaSO4 +
2HCl
+
NaOH ----------->
NaCl
+
NaCl +
AgNO3 ----------->
NaNO3 +
AgCl
HCl
+
H2O ( HOH)
V. Combustion- ( Also called”Burning”) A Reaction between a Hydrocarbon and oxygen gas to
form carbon dioxide and water.
General Forms:
Hydrocarbon + oxygen gas ------->
CO 2
+ H2O
Examples:
CH4
+
C9H18 +
2O2
----------->
CO2
+
2H2O
27/2 O2
---------->
9CO2
+
9H2O
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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ACTIVITY SERIES
LITHIUM
POTASSIUM
CALCIUM SODIUM
MAGNESIUM
ALUMINUM
ZINC
CHROMIUM
IRON
NICKEL
TIN
LEAD
HYDROGEN
COPPER
MERCURY
SILVER
PLATINUM
GOLD
HALOGENS
FLUORINE > CHLORINE> BROMINE > IODINE
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Worksheet FF CP Chem
Name ___________________
I. Balance the following Equations.
CH4
H2 O

HNO3
KI 
KBr
C
2.
N2O5 +
3.
Br2
+
4.
H2
+
Fe3O4

Fe
5.
Al
+
AgNO3

Al(NO3)3 +
6.
Al2(SO4)3 +
7.
FeCl3
8.
Ni(ClO3)2
9.
I2
+
NaCl

10.
H2
+
Br2

11.
H2SO4
12.
SrSO4
+

1.
H2
+
Sr(NO3)2 +
13.
P4
14.
Hg(OH)2
15.
C7H16 +
16.
S8
+
Ca(OH)2  Al(OH)3 +
H2S

FeCl2 +

I2
+

HCl
+
S
O2
+
Cl2
HBr
HOH
+
SO3

K2SO4
+
Ag
CaSO4
NiCl2 +
NaI
H2 O
O2

+
P2O5

HgO
+
O2

CO2
+
+
O2

KNO3
HOH
H2 O
SO2
II. Also, for each of the 16 questions above, identify the type of reaction as either Composition,
combustion Decomposition, single or double replacement, .
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Worksheet GG CP CHEM
Equations- writing & balancing,
Name________________
I. Priority #1- Write the correct formulas for each Reactant and Product. Then balance each
equation. Then try to classify the reactions as either composition, decomposition, single or double
replacement, or combustion.
1. Sodium
2. Calcium
+

Iodine
+
oxygen
 Calcium oxide
3. Nickel chlorate 
Nickel chloride + Oxygen
4. Barium Carbonate 
Barium oxide
5. Aluminum +
hydrochloric acid 
+ Chlorine 
6. Potassium Iodide
7. Copper II hydoxide + acetic acid
8. Hydrogen + Chlorine
9. Zinc hydroxide
10. Iron
+ Carbon dioxide


Potassium Chloride + Iodine

Copper II Acetate +
Water
Hydrogen chloride
Zinc Oxide
+ Copper II Nitrate
11. Manganese sulfate
loride
+ Water

+ ammonium sulfide
Iron II Nitrate

+ Copper
Manganese Sulfide
+ Ammonium Sulfate
12. sodium hydrogen carbonate ---> sodium carbonate + water + carbon dioxide
13. sodium + iodine ---> sodium iodide
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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14. barium chloride + potassium sulfate ---> barium sulfate + potassium chloride
15. calcium hydroxide + ammonium sulfate ---> calcium sulfate + water + ammonia
16. silver oxide ---> silver + oxygen
17. zinc + hydrochloric acid ---> hydrogen + zinc chloride
18. sodium hydroxide + sulfuric acid ---> sodium sulfate + water
19.
C2H2 +
20. Zinc
O2 --->
+ Sulfur
CO2 +
H2O

21. Potassium Chloride + Silver Nitrate
22. Calcium Oxide
+ water


Silver chloride + Potassium Nitrate
calcium hydroxide
23. Sodium hydroxide + hydrochloric acid  Sodium chloride + water

24. Aluminum + Iron III Oxide
25. Nitric Acid

water +
27. Hydrogen +
28. Zinc
Bromine
+
Nitrogen dioxide

26. Silver Nitrate + Nickel

sulfuric acid
Aluminum Oxide +
Iron
+
oxygen
Nickel Nitrate + Silver
Hydrogen Bromide

Zinc sulfate
+ hydrogen
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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Worksheet HH CP CHEM
Equations- writing & balancing.
Name_______________
I. Try to classify the reactions as either composition, decomposition, single or double replacement, or
combustion. Then balance each equation.
__________1.
Mg +
O2 --->
MgO
__________12.
Na2SO3 +
__________2.
Fe +
O2 --->
Fe2O3
__________13.
K2CO3 --->
K2O
__________3.
H2O +
SO2 --->
__________14.
NaOH --->
Na2O +
__________4.
Fe +
H2O --->
__________15.
Mg(ClO3)2 --->
H2SO3
Fe(OH)2
__________5.
Al +
Pb(NO3)2 --->
__________6.
Ca +
AgNO3 --->
__________7.
K+
H2O --->
__________8.
LiCl +
__________9.
Ca(OH)2 +
__________10.
KOH +
__________11.
Al(NO3)3 +
I2 --->
Al(NO3)3 +
Ca(NO3)2 +
KOH
LiI
HCl --->
H3PO4 --->
H2SO4 --->
+
+
Pb __________16.
Ag
__________17.
HOH
__________19.
+
K3PO4
Li2CO3 --->
HOH ________20. KClO3 --->
+
Al2(SO4)3
HOH
+
_________21.
HNO3
H2
+
+
O2
Ca(OH)2
Li2O
+
H2
+
H2SO3
HOH
H2SO4 --->
KCl
+
CO2
MgCl2
HCl --->
KOH +
NaCl
+
H2O --->
__________18. Zn +
Cl2
Ca Cl2
CaO +
HCl --->
CO2
+
ZnCl2
K2SO4
+
HOH
O2
F2 
HF
__________22. Na + Cl2 --> NaCl
II. Write the formulas, then balance each equation
__________23. hydrochloric acid + aluminum oxide --->
__________24. ammonium hydroxide + nitric acid --->
__________25. silver nitrate + hydrogen sulfide --->
Water + Aluminum chloride
Ammonium Nitrate
Silver Sulfide +
__________26. sodium hydroxide + phosphoric acid ---> water
+
+
Water
Nitric acid
Sodium Phosphate
__________27. mercury I nitrate + iron III chloride ---> Mercury I Chloride
+ Iron III Nitrate
__________28. C2H2 + oxygen --->
__________29. sulfur trioxide + water ---> Sulfuric acid
__________30. tin + phosphoric acid ---> Tin II Phosphate + Hydrogen
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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__________31. Mercury II hydroxide + potassium sulfate --->
Mercury II Sulfate
+
Potassium Hydroxide
__________32. siver nitrate + zinc chloride ---> Silver Chloride + Zinc Nitrate
__________33. oxygen + C7H16 --->
__________34. iron (III) + oxygen gas --->
Iron III Oxide
__________35. potassium chlorate ---> Potassium Chloride
__________36. iron + copper II chloride --->
+ oxygen
Copper + Iron III Chloride
__________37. magnesium + Iron III chloride --->
Magnesium Chloride + Iron
__________38. copper + magnesium chloride --->
Magnesium
__________39. calcium + hydrochloric acid --->
+
Copper II Chloride
Calcium Chloride
+ Hydrogen
__________40. cobalt + oxygen ---> Cobalt III Oxide
__________41. nickel sulfate + lithium phosphate --->
Lithium Sulfate +
__________42. sulfuric acid + sodium hydroxide --->
Water
Nickel Phosphate
+ Sodium Sulfate
__________43. C8H18 + oxygen --->
__________44. Iron II oxide + aluminum --->
Aluminum Oxide
+ Iron
__________45. calcium carbonate + hydrochloric acid --->
Calcium Chloride
__________46. barium hydroxide --->
water
__________47. zinc carbonate --->
Barium Oxide
Zinc oxide
+
+
+
Carbonic acid
Carbon dioxide
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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WORKSHEET II CP Chem
NAME_____________
(Optional)
I. CLASSIFY EACH OF THE REACTIONS BELOW( COMP, DECOMP, ETC) . WRITE THE
CORRECT FORMULAS AND BALANCE EACH EQUATION.
1.
Potassium + Bromine  Potassium Bromide
2.
Iron II Chlorate
3.
Aluminum
4.
C3H8 +
5.
Silver Nitrate
6.
Hydrogen
7.
Nickel Carbonate Nickel oxide + Carbon dioxide
8.
Iron II Sulfate +
Ammonium Sulfide Iron II Sulfide +
9.
Copper II Bromide
+
Acetic acid  Copper II Acetate + Hydrogen Bromide
10.
Barium Chloride
+
Sodium Sulfate
11
Manganese Hydroxide Manganese Oxide
12.
Calcium
+
Hydrochloric acid
13.
Iron II Sulfide +
hydrochloric acid
14.
Magnesium
+
Nitric Acid
15.
Potassium
+ HOH
16.
Lithium Iodide
+
17.
Silver
Sulfur
18.
Sodium Chlorate
19.
Zinc
20.
C6H12 +
 Iron II Chloride
+
Sulfuric Acid
oxygen
Aluminum Sulfate
+
hydrogen
oxygen
+
+
+
+
+
Zinc Chloride Silver Chloride +
Sulfur
Zinc Nitrate
Hydrogen sulfide
Barium Sulfate
Ammonium Sulfate
+
Sodium Chloride
+ water
Calcium Chloride + Hydrogen
Iron II Chloride + Hydrogen Sulfide
Magnesium Nitrate
Potassium Hydroxide
Chlorine gas
+ Hydrogen
+ Hydrogen
Lithium Chloride + Iodine
Silver Sulfide
Sodium Chloride + Oxygen
Lead IV Acetate
Zinc acetate + Lead
oxygen
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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21.
Lead II Acetete
22.
Aluminum Bromide +
23.
Magnesium Carbonate Magnesium Oxide
24.
Barium Bromate
25
Aluminum
+
Oxygen Aluminum oxide
26.
Sodium
+
HOH
27.
Calcium Carbonate
28.
Hydrogen peroxide Hydrogen
29.
Beryllium Hydroxide +
30.
Sodium Sulfite
31.
C4H10 +
oxygen
32.
Magnesium
+
33.
Iron III hydroxide Iron III Oxide
34.
CH4
35.
Calcium Nitride
+
Fluorine Calcium Fluoride + Nitrogen
36.
Barium Phosphate
+
Sodium perchlorate
37.
Nickel oxide +
Fluorine
38.
Gold chloride Gold
39.
cobalt II Iodide
+
40.
Zinc oxide
+
carbon dioxide
41.
Zinc oxide
+
water
42
Zinc chloride +
43.
water +
+
+
Hydrogen Sulfide
Fluorine
Lead II Sulfide + Acetic acid
Aluminum Fluoride + Bromine
+ Carbon dioxide
+ Potassium permanganate Barium permanganate + potassium Bromate
+
+
Sodium Hydroxide
+ Hydrogen
Hydrochloric acid Calcium Chloride + Carbonic acid
+ oxygen
Nitric acid
Beryllium Nitrate
Potassium Dichromate
+ Water
Potassium Sulfite + Sodium Dichromate
oxygen Magnesium Oxide
+ water
oxygen
Barium Perchlorate + Sodium Phosphate
Nickel Fluoride +
Oxygen
+ Chlorine
Sodium Sulfide
Cobalt II Sulfide + Sodium Iodide
Zinc Carbonate
Zinc hydroxide
oxygen Zinc Chlorate
carbon dioxide
Carbonic Acid
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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UNIT 8 NOTES- CP CHEM - STOICHIOMETRY
STOICHIOMETRYis the study of quantitative relationships between the amounts of reactants and
amounts of products, and energy used, in a chemical reaction(equation). Since matter cannot be created or
destroyed in a Chemical reaction, the total mass of products must be equal to the total mass of reactants.
For example, In the reaction below……
2H2
+
O2

2H20
+
energy
If 4 grams of Hydrogen gas react with 64 grams of oxygen gas, then the amount of water forming as a
product should be 68 grams. Even though atoms have been re-arranged to form new
molecules/compounds(products), there is no loss of mass of starting materials(reactants).
Also note that the mass ratios are 4: 64 : 68
This is not the same as the ratio of the coefficients used to balance the equation!!!!!
The coefficients tell us the number of molecules (or atoms)…..NOT the MASSES.
Since coefficients represent a quantity of atoms, they can also be used to represent MOLES,( and even
volumes of gases)…..so therefore…..
2 molecules H2
+
1 molecule O2  2 molecules H20
OR
2 moles H2
+
1 mole O2

2 moles H20
This reaction always takes place in a mole ratio of 2 : 1 : 2 no matter how many molecules. So , In
order to determine the amount of product manufactured, we will always need to express any “GIVEN”
information in “MOLES.
PARALLEL SITUATION
Suppose you work at “Meat Truck Company” on the assembly line. Here is a recipe for the
manufacturing of Meat’s best vehicle.
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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1 engine + 8 tires + 3 glass panes + 1 body 
1 truck
If you were asked: How many more cars could be made if 28 tires are remaining in the warehouse?
Since the numbers in the equation represent a quantity ratio, you can use them to determine your answer.
First: setup the factor label grid w/given and desired info.
28 tires
=
? cars
Second : use the equation to find the theoretical ratio of tires : cars
8 tires
or
1 truck
1 truck
8 tires
Third : Plug in the correct conversion factor and calculate
28 tires
1truck =
3.50 trucks = 3 trucks
8 tires
Lets try a REAL sample problem .
USE THIS PAGE AS A REFERENCE FOE ALL STOICHIOMETRY PROBLEMS
STOICHIOMETRY Sample Problem #1
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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How many grams of water vapor can be manufactured when 20.0 grams of Hydrogen gas reacts with (excess) oxygen .
First: Write and balance a chemical equation that represents the process. Interpret the equation in terms of moles
2H2
+
O2

2H20 + energy
Second: Since 20.0 g is a mass, convert it to moles ( a quantity) using the formula weight of hydrogen gas.
20.0 g H2
1 mole H2
2.0 g H2
Third : Use the equation to establish the ratio between Hydrogen gas (Given) and water vapor (the desired). Place
this conversion into the grid.
20.0 g H2
1 mole H2
2.0 g H2
2 mole H2O
2 mole H2
Fourth: Usually the mass(or volume ) is needed for the second substance , so convert moles of water back to grams
of water now using the formula weight of water .
20.0 g H2
1 mole H2
2.0 g H2
2 mole H2O
2 mole H2
18.0 g H2O
1 mole H2O
= 180.g H2O
STOICHIOMETRY Sample Problem #2
Ammonium Nitrate can be decomposed into water vapor and dinitrogen oxide. How many grams of Ammonium Nitrate
is needed to produce 180.0 g water ?
First: Write and balance a chemical equation that represents the process. Interpret the equation in terms of moles.
NH4NO3 (s) 
N2O (g)
+
2H20 (g)
Second: Since 180.0 g is a mass, convert it to moles ( a quantity) using the formula weight of water.
180.0 g H2O
1 mole H2O
18.0 g H2O
Third : Use the equation to establish the ratio between ammonium nitrate (the desired) and water vapor (the given).
Place this conversion into the grid.
180.0 g H2O
1 mole H2O
18.0 g H2O
1 mole NH4NO3
2 mole H2O
Fourth: Usually the mass(or volume ) is needed for the second substance , so convert moles of ammonium nitrate
back to grams using the formula weight of ammonium nitrate .
180.0 g H2O
1 mole H2O
18.0 g H2O
Worksheet JJ CP Chem
Stoich for starters
1 mole NH4NO
2 mole H2O
80.0 g NH4NO3 = 400.0 g NH4NO3
1 mole NH4NO3
Name_____________
I. Complete each of the problems below using the factor label method.
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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1. How many grams of chlorine gas are required to make 730 grams of Hydrogen chloride, assuming
enough H2 gas is present ?
H2
+
Cl2
 HCl
2. How many moles of oxygen gas are required to produce 242 grams of Magnesium Oxide in this
reaction ?:
Mg
+

O2
MgO
3. Silver reacts with Nitric acid as indicated by the following equation:
Ag
+
HNO3 
NO2
+
AgNO3+
HOH
What minimum amount of HNO3 would be required to react with 5.00 grams of silver ?
4. How many grams of oxygen are required to oxidize 140 grams of iron to Iron III Oxide?
Fe
b)
+
O2

Fe2O3
How many grams of Fe2O3 are produced ?
5. How many grams of water are produced when 136 grams of ammonia are burned in excess oxygen ?
The reaction is :
NH3
+
O2

Worksheet KK CP Chem
Stoichiometry
N2
+
H2O
Name_______________
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
20
I. Complete each of the problems below using the factor label method. Please complete in
your notebook.
1. If 120 grams of Sodium carbonate react with calcium hydroxide, how many grams of sodium hydroxide
are formed?
Na2CO3
+
Ca (OH )2
 NaOH +
CaCO3
2. If 90.0 grams of Barium chloride react with sulfuric acid, how many grams of Barium Sulfate are
produced?
BaCl2 +
H2SO4 
BaSO4 +
HCl
3. If 500.0 grams of Potassium Iodide react with lead II acetate, what weight of lead II iodide is formed?
KI + Pb(C2H3O2)2  PbI2 + KC2H3O2
4. When 80.0 grams of calcium chloride react with silver nitrate, how many grams of silver chloride are
produced ? Write the equation(there is one other product!).
5. How many grams of oxygen are produced by heating 400.0 grams of potassium chlorate ?.
KClO3  KCl + O2
6. If 600.0 grams calcium hydroxide react with nitric acid, how many grams of nitric acid are needed to
completely react with the calcium hydroxide ?
Ca(OH)2 + HNO3  HOH
+ Ca(NO3)2
7. When 50.0 grams of Mg react with AgNO3 in solution, how many grams of Ag are prepared ?
+ AgNO3 
Ag + Mg(NO3)2
Mg
8. How many grams of calcium carbonate are required in the preparation of 50.00 grams of calcium oxide
?
CaCO3
 CaO + CO2
9. Determine how much carbon dioxide and water will be produced from the combustion of 104 g of C2H2.
C2H2 + O2 
CO2 + H2O
10. How many moles of aluminum iodide are produced in the reaction between iodine and 108 g of Al.
11. What mass of sodium hydroxide is needed to produce 16.0 g of magnesium hydroxide if it reacts with
magnesium chloride?
NaOH + MgCl2  NaCl + Mg(OH)2
EXCESS REAGENT(Reactant)-LIMITING REAGENT(Reactant)
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The Limiting reagent limits the extent of a chemical reaction , and thus limits the amount of product that
can be formed in a chemical reaction. The Excess reagent is leftover, meaning that it remains in the
reaction vessel after the reaction has come to a stop .
Let’s go back to our analogy at “Meat Truck Company”. Here’s the recipe for the manufacturing of
Meat’s best vehicle.
1 engine + 8 tires + 3 glass panes + 1 body 
1 truck
If you were asked: How many trucks could be made from 52 tires and 66 glass panes?
Since the numbers in the equation represent a quantity ratio, you can use them to determine your answer.
Setup the factor label grid w/given and desired info for each reactant.
52 tires
1truck
8 tires
=
66 glass panes 1car
=
3 glass panes
6.5 trucks
22 trucks
Since the 52 tires can only produce 6 trucks, it is considered to be the limiting reagent. Once the six
trucks are manufactured, the reaction(assembly line) stops. No more trucks can be made. At this point
there are still enough glass panes left to make 16 more trucks. The glass panes are considered to be excess
reactant. See if you can determine “How many glass panes remain?”
Plug in the correct conversion factor and calculate
52 tires
66 panes in stock
3 panes
8 tires
= 19.5 panes used
- 19 panes used = 47 leftover panes
Sample #3: Excess- limiting reagent (Treuer’s Method)
In the production of disulfur dichloride, molten sulfur reacts with chlorine gas according to this
equation:
S8
+ 4Cl2 
4S2Cl2 (l)
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22
If 200.0 g Sulfur reacts with 100.0 g chlorine, what mass of disulfur dichloride is produced ?
Step 1 : Determine how many grams of disulfur dichloride is produced by each reactant.
200.0 g S8
1 mole S8
256.5 g S8
4 moles S2Cl2
1 mole S8
135.0 g S2Cl2 _ = 421 g S2Cl2
1 mole S2Cl2
100.0 g Cl2
1mole Cl2
71.0 g Cl2
4 moles S2Cl2 135.0 g S2Cl2 _ = 190.4 g S2Cl2
4 moles Cl2
1 mole S2Cl2
THE REACTANT THAT PRODUCES THE LESSER MOLE/grams AMOUNT OF
PRODUCT(S2Cl2) IS CONSIDERED TO BE THE LIMITING REACTANT (REAGENT)
Cl2 is the “Limiting Reagent”
S8 is the “Excess Reagent
B) To determine the amount of excess reagent reacted (used up)
…….. start with the actual limiting reagent, chlorine. Treat the excess reagent as the “desired”
100.0 g Cl2
1mole Cl2
71.0 g Cl2
1 mol S8
256.5 g S8. = 90.42 g S8 used
4 mole Cl2 1 mole S8
C) To determine the amount of excess reagent unreacted, leftover, or unused…
Take the amount that was used, 90.42 g Sulfur reacted, and subtract this amount from the amount of excess
reagent that you had initially. Because 200.0 g of sulfur is available and only 90.42 g is needed the mass in
excess is….
200.0 g S8 - 90.42 g S8 = 109.6 g S8 in excess
WORHSHEET LL H Chem Name ___________________________
Excess/Limiting Stoich “Have an excess amount of fun with these”
1. How many grams of Al2O3 can be formed when a mixture of 0.36 mol of aluminum and 0.36 mol of
oxygen is ignited?
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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2. What mass of bromine could be produced from 2.10 g of NaBr and 9.42 g of H2SO4?
2NaBr + 2H2SO4 + MnO2 ----> Br2 + MnSO4 + 2H2O + Na2SO4
3. If 20.0 g of NaOH react with 30.0 g of H2SO4, how much sodium sulfate will be produced?
NaOH + H2SO4  Na2SO4 + HOH
4. If 5.00 g of copper metal react with 20.0 g of silver nitrate, how much silver will be produced?
5. How many grams of iron III oxide can be made from 25.5 g of iron and 25.0 g of oxygen? What
substance is the limiting agent?
6
Given :
CaCO3
+
2NaCl

Na2CO3
+
CaCl2
How Many grams of calcium chloride are formed when 1000 grams of NaCl react with 2000 grams
of calcium carbonate ?
7.
Given :
C12H22O11 + 12 O2  12 CO2
+ 11 H2O
34.2 grams of sugar reacts with 32.0 grams of oxygen gas. How many moles of water is produced ?
8. Iron burns in air(oxygen) to form a black iron oxide, Fe3O4.
56 grams of iron reacts with 64 grams of oxygen. How much iron oxide is produced ?
PERCENTAGE YIELD
Quite often in chemistry, chemical reactions do NOT always produce the amount of product that is
calculated (on paper). In the calculations that you have been practicing, the answer that you obtain
on paper is called the Theoretical Yield. The Theoretical Yield is the maximum amount of product
Mr Treuer CP Chemistry Notes Unit 7 & 8 Equations / Stoichiometry 2012-2013
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that can be produced from a given amount of reactant. Since a chemical reaction rarely produces
the theoretical yield (in the real world), the amount of product that is truly produced in an
experiment is called the Actual Yield.
One way to decide how efficient a chemical reaction is in producing the desired product is to
perform a calculation known as Percent Yield. Mathematically , percent(age) yield can be
expressed as :
Percent Yield = Actual Yield (from the experiment)
x 100%
Theoretical Yield (from stoichiometric calculations)
Sample Problem # 4 Percent yield
a) Determine the theoretical yield of water when 136 grams of ammonia is burned in excess oxygen ?
The reaction is :
4NH3 +
3O2

2N2
+
6H2O
The theoretical yield is :
136 g NH3
1 mole NH3
17.0 g NH3
6mole H2O
4mole NH3
18 g H2O
= 216 g H2O (Theo.)
1 mole H2O
b) If 168 grams of water is obtained, calculate the percent yield.
The percent yield is :
% Yield = Actual Yield
Theoretical Yield
% Yield = 168 g
216 g
% Yield =
x 100%
x 100%
77.8 %
Stoichiometry Lab :
Iron - Copper II Sulfate Reaction
Stoichiometry is a Calculation in Chemistry that deals with quantitative relationships between the reactants,
products , and energy in chemical reactions.
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In this experiment, you will let iron filings react with Copper II Sulfate solution. An excess of Copper II
Sulfate will be used so that all of the iron filings will be reacted. By weighing the amount of iron used and
the amount of new substance formed, you will be able to determine a quantitative relationship between
reactants and products for this reaction.
As you do this experiment, record your data carefully and neatly. Take special care to show the units as an
important part of each measurement. Express each measurement using the correct number of significant
figures as determined by the precision of the measuring instrument.
Before coming to lab, you should plan what you are to do.
DATA TABLE:
1
2
3
4
5
6
7
8
9
10
Mass of 250 ml beaker (g)
Mass of Beaker and CuSO4 (g)(optional)
Mass of CuSO4 used (g)
Mass of weighing dish (g)
Mass of weighing dish and Fe (g)(option)
Mass of Fe used (g)
Mass of filter paper (g)
Mass of filter paper and product (g)
Mass of product (g)
Procedure:
a) weigh a clean, empty 250 ml beaker to the nearest 0.01 gram. Record in table(1)
b) Tare the beaker, add Copper II Sulfate to the beaker until the balance indicates a reading of just over 10 grams.
Record the mass of Copper II Sulfate in the data table (3)
c) Add about 50 ml of de-ionized water to the copper II Sulfate crystals.
d) Place the beaker on a ring stand with wire gauze. Warm the solution until it just starts to boil. Be careful not to le
the solution boil over!! Remove the burner.
e) Weigh a plastic weighing dish. Record the mass in your data table (4)
f) Tare the plastic dish and add Fe filings so that the mass is somewhere between 1.50 and 2.10 g of Iron. Record the
exact mass of Iron in the data table (6).
g) Weigh a piece of filter paper on the balance. Record the mass in the data table. Fold the paper into “quarters”
and write your initials on the edge of the paper (in pencil). Set the filter paper up inside a funnel on a funnel stand.
h) While stirring the hot copper sulfate solution with a glass rod, add the iron filings SLOWLY AND A LITTLE
AT A TIME. After the reaction is complete, slowly pour the contents (beaker is HOT) of the reaction into the filter
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paper, insuring that all solids are collected by the paper. Rinse your beaker out with a de-ionized water wash bottle.
Be sure to get out any solids in the beaker by washing them into the filter paper.
i) After the final washing, the solid must be dried. Carefully remove the filter paper (don’t spill) and let dry on the
side of the room, either in the drying oven or by air drying overnite.
j) After the solid is dry (next lab day) weigh the solid and filter paper on the balance (8)
Calculations/Questions
1. Determine the mass (experimental result) of the product (show Calculation in lab report)
2. What do you think the product is ? Why ?
3. Balance this equation for the reaction: Fe(II) + Cu SO4 
4. Calculate the moles of copper that can be produced from the moles of iron and Copper II Sulfate used in the
experiment. In other words, show which reactant is the limiting and Excess reagent ?
5. Determine the grams (Theoretical result) of copper formed in the reaction.
6. Using the Actual mass of Copper (ques #1),and the theoretical mass (ques 6) determine the percent yield.
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