 1. Limits or determines the amount of product that can be formed
 2. The reagent that is not used up is therefore the excess reagent
 These types of problems require 2 sets of tracks. Quantities of both reagents will be given. Therefore, you need to find out which one is the limiting reagent.




Limiting Reagent Example problem
 How many grams of copper (I) Sulfide can be produced when 80.0 grams of Cu reacts with 25.0 grams of sulfur?
 2Cu + S --> Cu
2
S
 Pick a reactant and calculate how much of the other reactant is needed.
80.0g Cu 1mol Cu 1mol S 32.1g S
63.5g Cu 2mol Cu 1mol S = 20.2g S
So, 20.2 g of S is needed; 25.0g is supplied
Plenty of S; therefore, Cu is limiting reagent.
Use Cu to solve the problem
80.0g Cu 1mol Cu 1mol Cu
2
S 159.1g Cu
2
S
63.5g Cu 2mol Cu 1mol Cu
2
S
= 1.00x10
2 g Cu
2
S

Limiting Reagent Example Problem - Your Turn
 How many grams of hydrogen can be produced when
5.00g of Mg is added to 6.00 g of HCl?
 Mg + 2 HCl --> MgCl
2
+ H
2
 Pick a reactant and calculate how much of the other reactant is needed.
5.00g Mg 1mol Mg 2mol HCl 36.5g HCl
24.3g Mg 1 mol Mg 1mol HCl = 15.0g HCl
Need 15.0g HCl; have 6.00 g HCl
Not enough HCl; therefore, HCl is limiting reagent
Use HCl to solve the problem
6.00g HCl 1mol HCl 1mol H
2
2.0g H
2
36.5g HCl 2mol HCl 1mol H
2
= 0.164 g H
2

Limiting Reagent Example problem- Your Turn
 Acetylene (C
2
H
2
) will burn in the presence of oxygen. How many grams of water can be produced by the reaction of
2.40 mol of acetylene with 7.4 mol of oxygen?
 2 C
2
H
2
+ 5 O
2
--> 4 CO
2
+ 2 H
2
O
 Pick a reactant and calculate how much of the other reactant is needed
 2.40 mol C
2

H
2
5 mol O
2
2 mol C
2
H
2
 Need 6.00mol O
2
; have 7.4mol O
2
 Plenty of O
2
; so, C
2
H
2 is L.R.
= 6.00 mol O
2
 2.4mol C
2
H
2

2mol H
2
O 18.0g H
2mol C
2
H
2
2
O
1mol H
2
O2
 = 43.2 g H
2
O