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

• 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 2S 2M 6E ?
Cake 20 8 30 48 0 8 4 8
cups of sugar will make 4 cakes All other
ingredients make more The sugar runs out-
limiting ingredient
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3F 2S 2M 6E ?
Cake 20 8 30 48 0 8 4 4
x the recipe
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3F 2S 2M 6E ?
Cake 20 8 30 48 0 12 8
4 4 x 3F 12F
7
3F 2S 2M 6E ?
Cake 20 8 30 48 0 12 8 8
4 4 x 2M 8M
8
3F 2S 2M 6E ?
Cake 20 8 30 48 0 12 8 8 24
4 4 x 6E 24E
9
3F 2S 2M 6E ?
Cake 20 8 30 48 0 12 8 8 24
4 Subtract to get whats let over
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3F 2S 2M 6E ?
Cake 20 8 30 48 0 12 8 8 24
4 8 0 22 24 4 S is the limiting
ingredient All other ingredients are in excess
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Limiting Reactant Problems
12

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 12.0
• Pick one reactant O2 to run out and check to see
  if you
• have enough Ga.

3
2
4
4
3
13

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 16.0 12.0
• Cannot consume 16.0 moles - only have 15 moles
• - Ga must run out

3
2
4
4
3
14

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 15.0

3
2
4
3
4
15

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 15.0 11.25

3
2
4
3
2
3
4
16

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 15.0 11.25 7.50
• Subtract reactants and add products.

3
2
4
3
2
3
4
17

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 15.0 11.25 7.50
• Subtract reactants and add products.

3
2
4
3
2
3
4
18

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 15.0 11.25 7.50
• End 0.00 0.75 mole 7.50 mole
• Convert back to grams

3
2
4
19

• 15.0 mole Ga and 12.0 mole O2 react. Find
• the limiting reactant, the mass of excess
• reactant and product made.
• Ga O2 ? Ga2O3
• Initial 15.0 12.0 0
• Change 15.0 11.25 7.50
• End 0.00 0.75 x 32 g 7.50 x 187.4 g
• 1 mole 1 mole
• limiting 24.0 g 1.41 x 103 g

3
2
4
20

• 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
• 14.0 g 94.0 g

3
2
2
21

• 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
• 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
• C 0.5185 mole

3
2
2
22

• 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
• 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
• C 0.5185 mole

3
2
2
3
2
23

• 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
• 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
• C 0.5185 mole 0.7778 mole
• Not enough Br2

3
2
2
3
2
24

• 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
• 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
• C 0.5882 mole

3
2
2
2
3
25

• 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
• 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
• C 0.3921 mole 0.5882 mole
• Enough Al

3
2
2
2
3
26

• 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
• 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
• C 0.3921 mole 0.5882 mole

3
2
2
2
2
3
3
27

• 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
• 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
• C 0.3921 mole 0.5882 mole 0.3921 mole

3
2
2
2
2
3
3
28

• 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
• 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
• C 0.3921 mole 0.5882 mole 0.3921 mole
• E 0.1264 mole 0.0000 0.3921
• x 27.0 g x 266.7 g
• 1 mole 1 mole
• 3.41 g limiting 105 g

3
2
2
2
2
3
3
29

• 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
• 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
• C 0.2550 mole

3
2
1
6
30

• 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
• 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
• C 0.2550 mole

3
2
1
6
31

• 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
• 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
• C 0.2550 mole 0.3825 mole

3
2
1
6
32

• 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
• 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
• C 0.2550 mole 0.3825 mole

3
2
1
6
33

• 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
• 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
• C 0.2550 mole 0.3825 mole 0.1275 mole

3
2
1
6
34

• 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
• 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
• C 0.2550 mole 0.3825 mole 0.1275 mole

3
2
1
6
35

• 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
• 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
• C 0.2550 mole 0.3825 mole 0.1275 mole 0.765
  mole

3
2
1
6
36

• 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
• 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
• C 0.2550 mole 0.3825 mole 0.1275 mole 0.765
  mole
• E 0 mole 0.1903 mole 0.1275 mole 0.765
  mole

3
2
1
6
37

• 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
• 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
• C 0.2550 mole 0.3825 mole 0.1275 mole 0.765
  mole
• 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

3
2
1
6
38

• 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
• 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
• C 0.2550 mole 0.3825 mole 0.1275 mole 0.765
  mole
• 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

3
2
1
6