Limiting Reagents

Example

Your job is to attach arms on dolls in a doll factory. If you have 600 arms and 350 doll bodies, how many finished dolls can you make?

What is limiting your production – the arms or the dolls?

What do you have in excess?

•

Reagent

–reactant in a chem reaction •

Limiting Reagent

– the reactant that is completely consumed. This reactant

limits

the amount of product that is formed.

•

Excess Reagent

– the reactant that is not limiting the reaction. There will be leftovers of this reactant.

How to solve for limiting reagent 1. Write the balanced chemical equation 2. Determine the moles of each reactant } 3. Determine how many moles of product each reactant would make } using a mole ratio 4. The reactant that yields less product is the limiting reagent.

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reagent when 80.0g Cu reacts with 25.0g S?

1. Write the balanced chemical equation 2 Cu + S  Cu 2 S

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reagent when 80.0g Cu reacts with 25.0g S?

2. Determine the moles of each reactant.

80.0 g Cu x 1 mole Cu 63.55 g Cu = 1.26 mol Cu 25.0 g S x 1 mole S 32.06 g S = .780 mol S

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reagent when 80.0g Cu reacts with 25.0g S?

3. Determine how many moles of product each reactant would make using a mole ratio 1.26 mol Cu x 1 mol Cu 2 S 2 mol Cu = .630 mol Cu 2 S .780 mol S x 1 mol Cu 2 S 1 mol S = .780 mol Cu 2 S

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reagent when 80.0g Cu reacts with 25.0g S?

4. The reactant that yields less product is the limiting reagent.

Copper yields .630 mol Cu 2 S Sulfur yields .780 mol Cu 2 S So Copper is the limiting reagent

• What is the excess reagent?

Sulfur

• How many moles of Cu

2

S will be produced?

.630 mol Cu 2 S

Amount of Excess?

• In order to determine how much (mass) of excess reagents there are, you must figure out what mass of the excess reagent got used up…then subtract that from the initial mass.

Amount of Excess?

• Step 1 – Start with moles of product formed (from limiting reagent).

• Step 2 – Convert back to moles of excess reagent.

• Step 3 – Convert from moles to mass of excess reagent.

• Step 4 – Subtract this value from the original mass of excess reagent (given in problem)

• Hydrogen fluoride reacts with aluminum oxide to produce aluminum fluoride and water.

– If 60.0g of hydrogen fluoride react with 25.0g of aluminum oxide, how many grams of aluminum fluoride will be produced?

– Determine the mass of excess reactant remaining.