Limiting Reactants and ICE Charts

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

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?

4
3F 2S 2M 6E ?
1Cake
20 8 30 48 0
Initial
8
4
12
8
24
Change
8
0
22
24
4
End
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.
Ga O2 ?
Ga2O3
3
2
4
14.0
0
I
12.0
4
12.0
C
16.0
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.
Ga O2 ?
Ga2O3
3
2
4
14.0
0
I
12.0
3
2
10.5
C
14.0
7.00
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
excess
limiting
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.

Al Br2 ? AlBr3
3
2
2
14.0 g x 1 mole 94.0 g x 1 mole
27.0 g 159.8 g
I 0.5185 mole 0.5882 mole 0 mole
3
0.7778 mole
C 0.5185 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.

Al Br2 ? AlBr3
3
2
2
14.0 g x 1 mole 94.0 g x 1 mole
27.0 g 159.8 g
I 0.5185 mole 0.5882 mole 0 mole
2
2
0.5882 mole
0.3921 mole
C
0.3921 mole
3
3
E 0.1264 mole 0.0000 0.3921 mole
x 27.0 g x 266.7 g
1 mole 1 mole
3.41 g limiting 105 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.
• H3PO4 Ca(NO3)2
  ? Ca3(PO4)2 HNO3

3
2
1
6
25.0 g x 1 mole 94.0 g x 1
mole 98.03 g 164.1 g
I 0.2550 mole 0.5728 mole 0 mole 0 mole
3
1
6
C 0.2550 mole
0.3825 mole
0.1275 mole
0.765 mole
2
3
1
E 0 mole 0.1903 mole 0.1275 mole 0.765 mole

x 164.1 g
x 310.3 g x 63.01 g
1 mole 1 mole
1 mole
limiting 31.2 g
39.6 g 48.2 g