Transcript Limiting Reactant

Limiting Reactant
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)
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.
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 + 1 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)
Solving Limiting Reactant
Problems
1.
2.
3.
4.
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
5. The reactant that produces the least amount of
product is the limiting reactant. Use this amount
to do any further calculations
E.g. What mass of magnesium oxide is produced when 6.73 g
of magnesium reacts with 8.15 g of oxygen
MM
n
2 Mg(s) +
O2(g) 
2 MgO(s)
24.31 g/mol
32.00 g/mol
40.31 g/mol
= 6.73 g X 1 mol = 8.15 g X 1 mol
24.31 g
32.00 g
= 0.2769 mol
= 0.2547 mol
Using Mg:
= 0.2769 molMg X 2 molMgO
2 molMg
= 0.2769 molMgO
Using O2:
= 0.2547 molO2 X 2 molMgO
1 molO2
= 0.5094 molMgO
0.2769 mol < 0.5094 mol  Mg limits
m
6.73 g
8.15 g
= 0.2769 mol X 40.31 g
mol
= 11.2 g
Practice!
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Worksheet B (no limiting reactant)
Worksheet C
P. 231 #1-4
P. 235 #5-10