Limiting Reactants and
ICE Charts
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
1.
How many cakes can you make?
2.
Which ingredient ran out first and limited the number of cakes you
could make?
3.
What and how much of each ingredient is left over?
4.
What does this assignment have to do with chemistry?
+
6E
→ 1Cake
3F +
2S +
2M
Initial
20
8
30
48
0
Change
12
8
8
24
4
End
8
0
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
Limiting Reactant Problems
14.0 mole Ga and 12.0 mole O2 react. Find the limiting reactant, the
mass of excess reactant and product made.
4 Ga
+
3 O2
→
2 Ga2O3
0
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
14.0 mole Ga and 12.0 mole O2 react. Find the limiting reactant, the
mass of excess reactant and product made.
4 Ga
+
3 O2
→
2 Ga2O3
0
I
14.0
12.0
3
2
C
7.00
14.0
10.5
4
3
7.00 x 187.4 g
E
0
1.5 x 32.0 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
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
+
3 Br2
14.0 g x 1 mole
27.0 g
I
C
94.0 g x 1 mole
159.8 g
0.5185 mole
0.5185 mole
→ 2 AlBr3
0.5882 mole
3
2
0.7778 mole
0 mole
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
+
3 Br2
14.0 g x 1 mole
27.0 g
I
94.0 g x 1 mole
159.8 g
0.5185 mole
0.5882 mole
2
C
E
0.3921 mole
3
0.1264 mole x 27.0 g
1 mole
3.41 g
→ 2 AlBr3
0.5882 mole
0 mole
2
3
0.3921 mole
0.0000
0.3921 mole x 266.7 g
1 mole
limiting
105 g
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.
2 H3PO4
3 Ca(NO3)2
1 Ca3(PO4)2 +
+
→
25.0 g x 1 mole
98.03 g
I
E
94.0 g x 1 mole
164.1 g
0.2550 mole
C 0.2550 mole
0 mole
0.5728 mole
3
2
0.3825 mole
0.1903 mole
x 164.1 g
1 mole
limiting
6 HNO3
= 31.2 g
0 mole
1
3
0.1275 mole
0.1275 mole
x 310.3 g
1 mole
= 39.6 g
0 mole
6
1
0.765 mole
0.765 mole
x 63.01 g
1 mole
= 48.2 g