# CHAPTER 11 Stoichiometry 11.3 Limiting Reactants.

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CHAPTER 11 Stoichiometry 11.3 Limiting Reactants

Recipe: Suppose you want to make 2 ham & cheese sandwiches
4 slices of bread 4 slices of ham 2 slices of cheese

Recipe: 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 & cheese sandwich 4 slices of bread 4 slices of ham 2 slices of cheese

No, you are limited by the cheese!
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.

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. Excess reactant

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.

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g)

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) 3 questions: 1. What is the limiting reactant? 2. How much excess reactant is left over? 3. How much product is formed?

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) 4 CH4 6 H2O Mole ratio:

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) 4 CH4 6 H2O Mole ratio: Limiting reactant Excess reactant

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) 3 questions: 1. What is the limiting reactant? 2. How much excess reactant is left over? 3. How much product is formed?

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) Mole ratio: 4 CH4 6 H2O

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) Mole ratio: 4 CH4 6 H2O 4 H2O moles react with 4 CH4 moles, so 2 H2O moles remain

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) 3 questions: 1. What is the limiting reactant? 2. How much excess reactant is left over? 3. How much product is formed?

Consider mixing 4 moles of methane and 6 moles of water.
Balanced equation: CH4(g) + H2O(l) → 3H2(g) + CO(g) 4 CH4

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles Step 2: Use mole ratios to find the limiting reactant

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles available Fe2O3 available Al Step 2: Use mole ratios to find the limiting reactant

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles available Fe2O3 available Al Step 2: Use mole ratios to find the limiting reactant

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles available Fe2O3 available Al Step 2: Use mole ratios to find the limiting reactant needed to react with all Fe2O3 needed to react with all Al

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g Step 1: Convert masses to moles available Fe2O3 available Al Step 2: Use mole ratios to find the limiting reactant needed to react with all Fe2O3 needed to react with all Al

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 150 g 60 g available Fe2O3 There is not enough Fe2O3 available to react with all the Al, so Fe2O3 is the limiting reactant needed to react with all Al

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 2. How much product (Fe) is formed? 150 g 0.94 moles 60 g 2.22 moles Limiting reactant

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 2. How much product (Fe) is formed? 150 g 0.94 moles 60 g 2.22 moles Limiting reactant Step 1: Use the limiting reactant and the mole fraction to find moles of Fe

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 2. How much product (Fe) is formed? 150 g 0.94 moles 60 g 2.22 moles Limiting reactant 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

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 2. How much product (Fe) is formed? 150 g 0.94 moles 60 g 2.22 moles Limiting reactant 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

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 2. How much product (Fe) is formed? 150 g 0.94 moles 60 g 2.22 moles Limiting reactant 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

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
1. What is the limiting reactant? 2. How much product (Fe) is formed? 150 g 0.94 moles 60 g 2.22 moles Limiting reactant 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

The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)

The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)

The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)

To calculate the amount of product from the amount of reactant:
Excess reactant Limiting reactant To calculate the amount of product from the amount of reactant: