#### Transcript limiting reactant

```CHAPTER 11
Stoichiometry
11.3 Limiting
Reactants
Suppose you want to make 2 ham & cheese sandwiches
Recipe:
2 ham &
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4 slices of ham
2 slices of cheese
cheese
sandwich
11.3 Limiting Reactants
Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you have
4 slices of bread, 4 slices of ham, and 1 slice of cheese?
Recipe:
2 ham &
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4 slices of ham
2 slices of cheese
cheese
sandwich
11.3 Limiting Reactants
Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you have
4 slices of bread, 4 slices of ham, and 1 slice of cheese?
Limiting factor
No, you are limited by the cheese!
You can only get 1 ham & cheese sandwich.
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11.3 Limiting Reactants
For a chemical reaction:
Reactant A is in excess
so the reaction will stop when you
run out of reactant B.
Reactant B is the limiting
reactant.
The amount of product C will
depend on how much reactant B
is present.
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Excess reactant
11.3 Limiting Reactants
Limiting
reactant
Excess
reactant
limiting reactant: the reactant that “runs out” first in a
chemical reaction.
excess reactant: the reactant that is remaining after the
reaction is complete.
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11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
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→
3H2(g) + CO(g)
11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
3 questions:
→
3H2(g) + CO(g)
1. What is the limiting reactant?
2. How much excess reactant is left over?
3. How much product is formed?
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11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
4 CH4
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6 H2O
→
3H2(g) + CO(g)
Mole ratio:
1 molecule CH4
1 molecule H2O
11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
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4 CH4
6 H2O
Limiting
reactant
Excess
reactant
→
3H2(g) + CO(g)
Mole ratio:
1 molecule CH4
1 molecule H2O
11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
3 questions:
→
3H2(g) + CO(g)
1. What is the limiting reactant?
2. How much excess reactant is left over?
3. How much product is formed?
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11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
4 CH4
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6 H2O
→
3H2(g) + CO(g)
Mole ratio:
1 molecule CH4
1 molecule H2O
11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
4 CH4
6 H2O
→
3H2(g) + CO(g)
Mole ratio:
1 molecule CH4
1 molecule H2O
4 H2O moles react with 4 CH4 moles, so 2 H2O moles remain
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11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
3 questions:
→
3H2(g) + CO(g)
1. What is the limiting reactant?
2. How much excess reactant is left over?
3. How much product is formed?
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11.3 Limiting Reactants
Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l)
3 molecules H2
1 molecule CH4
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and
1 molecule CO
1 molecule CH4
4 CH4
→
3H2(g) + CO(g)
12 moles H2
and 4 moles CO
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
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1. What is the limiting reactant?
60 g
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
Step 1: Convert masses to moles
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11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
Step 1: Convert masses to moles
Step 2: Use mole ratios to find the limiting reactant
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11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
Step 1: Convert masses to moles
150 g Fe2O3 
60.0 g Al 
1 mole Fe2O3
 0.94 moles Fe 2O3
159.7 g Fe2O3
1 mole Al
 2.22 moles Al
26.98 g Al
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant
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11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
Step 1: Convert masses to moles
150 g Fe2O3 
60.0 g Al 
1 mole Fe2O3
 0.94 moles Fe 2O3
159.7 g Fe2O3
1 mole Al
 2.22 moles Al
26.98 g Al
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant
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11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
Step 1: Convert masses to moles
150 g Fe2O3 
60.0 g Al 
1 mole Fe2O3
 0.94 moles Fe 2O3
159.7 g Fe2O3
1 mole Al
 2.22 moles Al
26.98 g Al
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant
0.94 moles Fe2O3 
2.22 moles Al 
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2 moles Al
 1.88 moles Al
1 moles Fe2O3
1 mole Fe2O3
 1.11 moles Fe2O3
2 moles Al
needed to react with all Fe2O3
needed to react with all Al
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
Step 1: Convert masses to moles
150 g Fe2O3 
60.0 g Al 
1 mole Fe2O3
 0.94 moles Fe 2O3
159.7 g Fe2O3
1 mole Al
 2.22 moles Al
26.98 g Al
available Fe2O3
available Al
Step 2: Use mole ratios to find the limiting reactant
0.94 moles Fe2O3 
2.22 moles Al 
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2 moles Al
 1.88 moles Al
1 moles Fe2O3
1 mole Fe2O3
 1.11 moles Fe2O3
2 moles Al
needed to react with all Fe2O3
needed to react with all Al
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
1. What is the limiting reactant?
60 g
150 g Fe2O3 
60.0 g Al 
1 mole Fe2O3
 0.94 moles Fe 2O3
159.7 g Fe2O3
available Fe2O3
1 mole Al
 2.22 moles Al
26.98 g Al
There is not enough Fe2O3 available to react with all the Al,
so Fe2O3 is the limiting reactant
0.94 moles Fe2O3 
2.22 moles Al 
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2 moles Al
 1.88 moles Al
1 moles Fe2O3
1 mole Fe2O3
 1.11 moles Fe2O3
2 moles Al
needed to react with all Al
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
60 g
0.94 moles 2.22 moles
Limiting reactant
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1. What is the limiting reactant?
2. How much product (Fe) is
formed?
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
60 g
0.94 moles 2.22 moles
Limiting reactant
1. What is the limiting reactant?
2. How much product (Fe) is
formed?
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
‹#›
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
150 g
60 g
0.94 moles 2.22 moles
Limiting reactant
1. What is the limiting reactant?
2. How much product (Fe) is
formed?
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
Step 2: Convert moles of Fe to mass of Fe
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11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is
150 g
60 g
0.94 moles 2.22 moles
formed?
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
0.94 moles Fe2O3  available  
2 moles Fe
 1.88 moles Fe formed
1 mole Fe2O3
Step 2: Convert moles of Fe to mass of Fe
‹#›
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is
150 g
60 g
0.94 moles 2.22 moles
formed?
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
0.94 moles Fe2O3  available  
2 moles Fe
 1.88 moles Fe formed
1 mole Fe2O3
Step 2: Convert moles of Fe to mass of Fe
‹#›
11.3 Limiting Reactants
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant?
2. How much product (Fe) is
150 g
60 g
0.94 moles 2.22 moles
formed?
Limiting reactant
Step 1: Use the limiting reactant and the mole fraction to find moles of Fe
0.94 moles Fe2O3  available  
2 moles Fe
 1.88 moles Fe formed
1 mole Fe2O3
Step 2: Convert moles of Fe to mass of Fe
1.88 moles Fe 
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55.85 g Fe
 105 g Fe formed
1 mole Fe
11.3 Limiting Reactants
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11.3 Limiting Reactants
The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)
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11.3 Limiting Reactants
The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)
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11.3 Limiting Reactants
The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)
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11.3 Limiting Reactants
Limiting
reactant
Excess
reactant
To calculate the amount of product from the amount of reactant:
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11.3 Limiting Reactants
```