1. Limiting Reactant

2. Stoichiometry Refresher
Recall:
The mole ratio can be used to determine the quantity (moles) or mass of reactants consumed and products formed
However, usually there is one reactant which will be consumed entirely and the other reactant is left over (in excess)

3. Excess Reactant: The reactant that is NOT completely consumed in a chemical reaction (i.e. it is in excess)
Limiting Reactant: The reactant that is completely consumed in a chemical reaction. It limits how much product is formed.

4. BBQ Analogy You are having a BBQ and want to make hot dogs.
You buy 2 packs of wieners and 2 packs of buns
There are 10 wieners per pack and 8 buns per pack (Marketing ploy??)
How many hot dogs can you make?
Write the BCE
1 wiener bun  1 hot dog
You can make 16 hot dogs and there will be 4 wieners left over (excess reactant)
You are limited by the number of buns (limiting reactant)

5. Solving Limiting Reactant Problems
Write the BCE
Set up a stoich table
Convert reactants into moles
Use mole ratios from BCE to calculate the moles of product formed using each reactant
The reactant that produces the least amount of product is the limiting reactant. Use this amount to do any further calculations

6. E. g. What mass of magnesium oxide is produced when 6
E.g. What mass of magnesium oxide is produced when 6.73 g of magnesium reacts with g of oxygen
2 Mg(s) +
O2(g) 
2 MgO(s) MM
24.31 g/mol
32.00 g/mol
40.31 g/mol n
= 6.73 g X 1 mol
24.31 g
= mol
= 8.15 g X 1 mol
32.00 g
= mol
Using Mg:
= molMg X 2 molMgO
2 molMg
= molMgO
Using O2:
= molO2 X 2 molMgO
1 molO2
= molMgO
mol < mol  Mg limits m
6.73 g
8.15 g
= mol X g
mol
= 11.2 g

7. Practice! Worksheet B (no limiting reactant) Worksheet C P. 231 #1-4