Unit 8 Review- Stoichiometry
Name:___________________
Balancing Equations
A subscript is a small number that tells you how many atoms are in a compound.
A coefficient tells also tells us how many atoms or compounds there are, but in a different way.
Ca3N2 + NaCl  CaCl2 + Na3N
Begin by “balancing” one atom at a time:
1. First, let’s balance the calcium atoms by placing a three in front of CaCl2:
Ca3N2 + NaCl  3 CaCl2 + Na3N
2. Next, let’s balance the nitrogen atoms, by placing a 2 in front of Na3N:
Ca3N2 + NaCl  3 CaCl2 + 2 Na3N
3. Now we will balance the sodium atoms by placing a 6 in front of NaCl.
Ca3N2 + 6 NaCl  3 CaCl2 + 2 Na3N
4. Finally, examine the chlorine atoms and notice that they are already balanced.
5. Double check each atom to make sure there are equal numbers of each on both sides of the
equation.
When balancing equations you NEVER change subscripts. Only change the coefficients.
o _____ CaCl2 + _____ Li2O  _____ CaO + _____ LiCl
o _____ Al2S3 + _____ Cu  _____ CuS + _____ Al
o _____ Al2(CO3)3 + _____ MgCl2  _____ AlCl3 + _____ MgCO3
o _____ Ba3(PO4)2 + _____ Na2CO3  _____ Na3 PO4 + _____ BaCO3
o
Na2S + Be(OH)2 
o
BaF2 + NaNO3 
Mole Ratios:
The numbers of moles of each substance in a chemical equation are related by the ratio of the
coefficients of each substance.
Reactants
Products
1 C3H8 + 5 O2
Mole Ratios:
1
:
3 CO2 + 4 H2O

5 :
3
:
4
Mole – Mole Calculations:

If you are given a number of moles of a reactant or product, you can determine the number of moles of any
other reactant or product.
# of Moles (given)
Mole
Ratio
*How many moles of CO2 are produced if you fully combust 3 moles of C3H8 (propane)?
3 moles C3H8
3 moles CO2
= 9 moles of CO2
1 mole C3H8
 How many moles of oxygen (O2) are produced by the decomposition of 6 moles of KClO3?
Balanced Chemical Equation:
Mole ratio:
___moles KClO3
2KClO3
2
:
2KCl + 3O2
2
:
3
___moles O2
___mole KClO3
=
 How many moles of KClO3 are needed to produce 3.75 moles of KCl?
Mole – Mass Calculations:


If you are given the moles of any reactant or product you can find out the number of moles of any other
reactant or product.
Once you calculate the number of moles of any reactant or product, you can find the mass of that reactant or
product.
# of Moles (given)
# of moles
Mole
Ratio
= # of moles
Molar mass (g)
1 mole
= mass (g)
 What is the mass in grams of potassium chloride (KCl) if 1.25 moles of potassium chlorate (KClO3) fully
decomposes?
Balanced Chemical Equation:
Mole ratio:
___moles KClO3
___moles KCl
___mole KClO3
2KClO3
2
:
2KCl + 3O2
2
:
3
___ moles KCl
____mass (g) KCl
=_______
=____
1 mole KCl
 What is the mass of oxygen(O2) produced from decomposition of 4.25 moles of potassium chlorate (KClO3)?
 What is the mass of potassium chlorate (KClO3) in the above balanced chemical equation if the reaction has
produced 5 moles of KCl?
Mass – Mole Calculations:


If you are given the mass of any reactant or product, you can find the number of moles of that reactant or
product.
Once you have number of moles, you can find the number of moles of any other reactant or product.
Mass (given)
1 mole
molar mass (g)
Mole ratio:

= # of Moles
# of Moles
(calculated)
Mole
= # of Moles
Ratio
3 Na2SO4 + Ca3P2
2 Na3P + 3 CaSO4
3
:
1
: 2
:
3
In the above reaction, how many moles of CaSO4 are produced when 250 g of Ca3P2 react completely
with Na2SO4?
 Considering the above balanced chemical equation, if 65 g of Na3P are produced, how many moles of
Na2SO4 must have reacted?
Mass – Mass Calculations:




In a balanced chemical equation, if you know the mass of any reactant or product, you can calculate the mass of
any other reactant or product.
First you must calculate the number of moles of the given mass.
Next you must compare the number of moles to the mole ratio from the balanced equation
Finally, when you have calculated the number of moles of your desired reactant or product, you must convert
back to mass using grams/mole of that substance.
mass (given)
1 mole
= # of moles
molar mass (g)
# of moles
(calculated)
mole
Ratio
Mole ratio:
# of moles
= # of moles
2 C4H10 + 13 O2
2 : 13 :
molar mass (g)
1 mole
= mass (g)
8 CO2 + 10 H2O
8
: 10
Consider the above balanced chemical reaction:
 If 75.3 g of C4H10 react with plenty of O2, what is the mass in grams of CO2 and H2O that can be
produced?
 If you need to produce 325 g of H2O, how many grams of C4H10 must react with an excess of oxygen?
Limiting Reactants:
Limiting reactant- the substance in a chemical reaction that limits the amount of products that can be
produced.
Excess reactant – the substance in a chemical reaction that is in excess (left overs).
Percent Yield:
Percent Yield - is a measure of how much of a product you actually obtained in a reaction compared to the
amount that you could have obtained according to calculations.
Actual Yield -the amount of product you actually collected.
Theoretical Yield - is the amount of a product that we should be able to make.
percent yield 
actual yield
 100
theoretical yield
 In the above balanced chemical reaction, the theoretical yield of NaCl is 109g. In an actual experiment you were
only able to collect 84.0g of NaCl. Calculate your percent yield.
 When 24.5 g of CaCl2 reacted with plenty of AgNO3, 21.5 grams of Ca(NO3)2 were produced.
What was the percent yield of Ca(NO3)2? Hint: first balance the following equation, then
calculate your theoretical yield using the mole ratios!
___CaCl2 + __AgNO3
___Ca(NO3)2 +___ AgCl