1. Mathematics of Chemical Equations
By using “mole to mole” conversions and balanced equations, we can calculate the exact amounts of substances that will be used up or produced during a reaction.

2. What you’ll learn: Write mole ratios from a balanced chemical equation
Calculate the number of moles and the mass of a reactant or product when given the number of moles or the mass another reactant.
Identify the limiting reactant in a chemical reaction.
Determine the percent yield of a chemical reaction.

3. Basic Example
N2 + 3H2  2NH3
From this balanced equation, we know that for every 1 mole of N2 we need ___ moles of H2 to react in order for ___ moles NH3 to be produced.
How many moles of N2 would we need to react with 6 moles of H2? 3 2 1
moles N2
6 moles H2 x
= 2 moles of N2
moles H2 3

4. N2 + 3H2  2NH3
How many moles of Ammonia (NH3) are produced if you have an excess of Nitrogen gas, and a sample of 0.43 moles of Hydrogen?
moles NH3 2
= moles
of NH3
0.43 moles H2 x
moles H2 3

5. Mass – Mass Conversions
You MUST BALANCE the equation first!
NaClO3  NaCl O2
How many grams of O2 would be produced if the final mass of NaCl produced was g?
375.6 g NaCl 2 2 3
mole O2 3
32 Grams O2
mole NaCl 1 x x x 1
58.5 g NaCl
mole NaCl 2
mole O2
= g O2

6. Stoichiometric Conversions A Three Step Process
Step 1 Convert the value of the given substance to moles of that substance.
Step 2 Convert from moles of the given substance to moles of the unknown substance (use the mole ratio from the balanced equation).
Step 3 Convert from moles of the unknown to proper unit.

7. Limiting Reactants

8. Limiting Reactants
When calculating your theoretical products, it is important to factor in which reactant you will run out of first!
This is called the “Limiting Reactant”. The other reactant(s) is called the “Excess Reactant”.

9. Step 1: Convert both to moles. 10gO2 x 1 mole O2 = 0.31 moles O2
Ex: If you have 10grams of O2 and 10 Liters of H2, how many molecules of water can you create? 2H2 + O2  2H2O
Step 1: Convert both to moles.
10gO2 x 1 mole O2 = moles O2
g O2
10L H2 x 1 mole H2 = moles H2
L H2

10. Step 2: Complete a mole to mole conversion, from one reactant to the other reactant to find the theoretical need.
2H2 + O2  2H2O
0.31 moles O2 x 2 moles H2 = 0.62 moles H2
1 mole O NEEDED
YOU KNOW that you have 0.45 moles H2 from your given values. So, that means you do not have enough H2. It is therefore the Limiting Reactant.

11. IF YOU COMPLETED STEP 2 FINDING MOLES OF O2:
2H2 + O2  2H2O
0.45 moles H2 x 1 moles O2 = moles O2
2 mole H NEEDED
You would come to the same conclusion, because this shows that you have more than enough O2. (You have .31 moles)

12. Step 3: Decide from Step 2 which reactant is the L. R
Step 3: Decide from Step 2 which reactant is the L.R. and use its moles to calculate your answer. (MUST have a balanced Chemical Equation!)
2H2 + O2  2H2O
0.45mol H2 x 2 mol H2O x 6.02x1023 molecules H2O =
mol H mole H2O
2.71x1023 molecules H2O

13. Percent Yield
Most chemical reactions never succeed in producing the predicted amount of product.
So, the actual amount of product is less than expected, due to experimental error
This is generally due to experimental error such as evaporation, product left on filter paper etc.

14. Percent Yield
Percent yield is the ratio of actual yield to the theoretical yield expressed as a percent.