Limiting Reactant

• Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data.

Include: theoretical yield, experimental yield

Additional KEY TermsExcess reactant

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: Record what you HAVE

Step 2: Calculate what you NEEDPick one reactant and calculate how much of the other you will need.

Step 3: Identify the limiting reactant

Step 4: Use limiting reactant to determine the amount of product.

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)

1 mol Cl2

2 mol Na6.70 mol Na Cl2 (need)

Pick one reactant and calculate the other

HAVE

NEED

6.70 mol 3.20 mol

= 3.35 mol

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

1 mol Cl2

2 mol Na3.20 mol Cl2

Both calculations lead to the same conclusion:

Have too much Na and Don’t have enough Cl2

Pick one reactant and calculate the other

HAVE 6.70 mol 3.20 molNEED 3.35 mol

Na (need)= 6.40 mol

= 6.40 mol NaCl1 mol Cl2

2 mol NaCl3.20 mol Cl2

You could use your data to calculate exactly how much excess is left over:

6.70 mol - 6.40 mol = 0.30 mol Na excess

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

Na - excess reactant Cl2 - limiting reactant

HAVE 6.70 mol 3.20 molNEED 6.40 mol 3.35 mol-

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

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 mol N2

3 mol H2

1 mol H2

2.0 g H2

3.50 g H2 28.0 g N2

1 mole N2

= 16.3 g N2

HAVE 18.0 g 3.50 g

NEED 16.3 g

3 mol H2

1 mol N2

1 mol N2

28.0 g N2

18.0 g N2 2.02 g H2

1 mole H2

= 3.90 g H2

3.90 g

18.0 g – 16.3 g = 1.70g N2 left

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

2 mol NH3

3 mol H2

1 mol H2

2.0 g H2

3.50 g H2 17.0 g NH3

1 mol NH3

= 19.8 g NH3

N2 - excess reactant H2 - limiting reactant

HAVE 18.0 g 3.50 gNEED 16.3 g 3.90 g

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

HAVE 75.0 g 150.0 L

NEED 58.9 g 191 L

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

= 191 L O2

5 mol O2

1 mol C3H8

1 mol C3H8

44 g C3H8

75 g C3H8 22.4 L O2

1 mol O2

= 58.9 g C3H8

1 mol C3H8

5 mol O2

1 mol O2

22.4 L O2

150 L O2 44 g C3H8

1 mol C3H8

75.0 g – 58.9 g = 16.1 g C3H8 left

C3H8 - excess reactant O2 - limiting reactant

HAVE 75.0 g 150.0 L

NEED 58.9 g 191 L

C3H8 + 5 O2 3 CO2 + 4 H2O

= 90 L CO23 mol CO2

5 mol O2

1 mol O2

22.4 L O2

150 L O2 22.4 L CO2

1 mol CO2

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

When solving limiting reactant problems:

1. Balance the chemical equation first2. Find the limiting reactant3. Use limiting reactant to determine the product4. Calculate the excess

CAN YOU / HAVE YOU?

• Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data.

Include: theoretical yield, experimental yield

Additional KEY TermsExcess reactant