Limiting Reactant

· Determine which reactant is left over in a reaction.

· Identify the limiting reactant and calculate the mass of the product.

· Calculate the amount of excess reactants.

How many BLT Sandwiches can you make with 3 Tomatoes, 2 Kg of Bacon, 1 head of lettuce and 4 slices of bread?

2

Limiting Reactant - determines the amount of product that can be formed in a reaction.

2 moles + 3 moles

Reactants remaining are called the excess reactants.

H

H

H

NH H

N

N

H H

H H

H

H

H

N

N2 (g) + 3 H2 (g) 2 NH3 (g)

N

N

Limiting Reactant Problems

Step 1: Determine which reactant will be used up first – pick one and calculate the other. (limiting reactant)

Step 2: Use the limiting reactant to determine the amount of product.

Step 3: Calculate the amount of reactant left over from the moles of limiting reactant used.

How much NaCl is produced and what is left over when 6.70 mol of Na react with 3.20 mol of Cl2?

2 Na (s) + Cl2 (g) 2 NaCl (s)

Cl2 - limiting reactant, Na - excess reactant

1 mol Cl2

2 mol Na6.70 mol Na

= 3.35 mol Cl2

= 6.40 mol NaCl1 mol Cl2

2 mol NaCl3.20 mol Cl2

= 6.40 mol Na1 mol Cl2

2 mol Na3.20 mol Cl2

6.70 mol - 6.40 mol = 0.30 mol Na excess

2 Na (s) + Cl2 (g) 2 NaCl (s)

…when 6.70 mol of Na react with 3.20 mol of Cl2?

Cl2 - limiting reactant, Na - excess reactant

N2 (g) + 3 H2 (g) 2 NH3 (g)

H2 - limiting reactant, N2 - excess reactant

How many grams of ammonia can be made from 3.50 g of H2 gas and 18.0 g of nitrogen gas? What’s left?

1 mole N2

3 mole H2

1 mole H2

2.0 g H2

3.50 g H2 28.0 g N2

1 mole N2

= 16.3 g N2

18.0 g – 16.3 g = 1.68g N2 left

N2 (g) + 3 H2 (g) 2 NH3 (g)

2 mole NH3

3 mole H2

1 mole H2

2.0 g H2

3.50 g H2 17.0 g NH3

1 mole NH3

= 19.8 g NH3

C3H8 (g) + O2 (g) CO2 (g) + H2O (g)

O2 - limiting reactant, C3H8- excess reactant

What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and 150.0 L of oxygen at STP? 3

= 191 L O2

5 mole O2

1 mole C3H8

1 mole C3H8

44.0 g C3H8

75.0 g C3H8 22.4 L O2

1 mole O2

3 4 5

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

= 58.9 g C3H8

= 90 L CO2

3 mole CO2

5 mole O2

1 mole O2

22.4 L O2

150 L O2 22.4 L CO2

1 mole CO2

1 mole C3H8

5 mole O2

1 mole O2

22.4 L O2

150 L O2 44.0 g C3H8

1 mole C3H8

75.0 g – 58.9 g = 16.1 g C3H8 left

· The limiting reactant is completely consumed. · The excess reactant is NOT used up.

When solving limiting reactant problems:

1. Balance the chemical equation first.2. Find the limiting reactant.3. Use the limiting reactant to determine the moles of

the product formed.4. Calculate the excess from the limiting reactant.