1. Limiting Reactants and ICE Charts

2. Chemistry Cake
You have 20 cups of flour, 8 cups of sugar, 30 litres of
milk and 48 eggs in your kitchen. The recipe for
chemistry cake is:
3 cups of flour
2 cups of sugar
2 litres of milk
+ 6 eggs
= 1 chemistry cake

3. How many cakes can you make?
Which ingredient ran out first and limited the number of cakes you could make?
What and how much of each ingredient is left over?
What does this assignment have to do with chemistry?

4. 3F S M E → 1Cake
Initial
Change 12 8 8 24 4
End 8 22 24 4
The sugar runs out- limiting ingredient
8 cups of sugar will make 4 cakes- so multiple the recipe by 4
All other ingredients are in excess

5. Limiting Reactant Problems

6. 14.0 mole Ga and 12.0 mole O2 react. Find the limiting reactant, the
mass of excess reactant and product made. 4
Ga O → Ga2O3 3 2 I
14.0
12.0 4 C
16.0
12.0 3
Guess that one of the reactants runs out- then check
Cannot consume 16.0 moles – we only have 14 moles

7. 14.0 mole Ga and 12.0 mole O2 react. Find the limiting reactant, the
mass of excess reactant and product made. 4
Ga O → Ga2O3 3 2 I
14.0
12.0 3 2 C
14.0
10.5
7.00 4 3
7.00
x g E
1.5
x g
1 mole
1 mole
48. g
1.31 x 103 g
limiting
excess
product
The other reactant must run out- calculate the change
Subtract to get what is left- Convert back to grams

8. 14.0 g of Al reacts with 94.0 g of Br2. Find the limiting reactant,
the mass of the excess reactant and product. 2
Al Br → AlBr3 3 2
14.0 g x 1 mole g x 1 mole
27.0 g g
I mole mole 0 mole 3
C mole
mole 2

9. 14.0 g of Al reacts with 94.0 g of Br2. Find the limiting reactant,
the mass of the excess reactant and product. 2
Al Br → AlBr3 3 2
14.0 g x 1 mole g x 1 mole
27.0 g g
I mole mole 0 mole 2 2 C
mole
mole
mole 3 3
x g x g
1 mole mole
E mole mole
3.41 g limiting g

10. 25.0 g of H3PO4 reacts with 94.0 g of Ca(NO3)2. Find the limiting
reactant, the mass of the excess reactant and product.
H3PO Ca(NO3) → Ca3(PO4)2 + HNO3 2 3 1 6
25.0 g x 1 mole g x 1 mole
98.03 g g
I mole mole 0 mole mole 3 1 6
C mole
mole
mole
0.765 mole 2 3 1
E 0 mole mole mole mole
x g x g x g
1 mole mole mole
limiting = g = g = g