Limiting Reactants and Excess

• What is the Limiting Reagent (Reactant)?It is the substance in a chemical reaction that runs out first.

• The limiting reactant (reagent) determines how much product you can make.

• If you are given amounts of more than one reactant, determine how much product you can make with each of them. Whichever produces the LEAST amount of product is your limiting reagent.

Practice ProblemsConsider the reaction:

2 Al + 3 I2 2 AlI3

Determine the limiting reagent of the product, aluminum iodide, if

one starts with:

a). 1.20 moles of Al and 2.40 moles of iodine

A. Al

B. I2

• 1.20 moles of Al 2 moles AlI3 1.20 moles AlI3

• 2 moles Al

• 2.40 moles I2 2 moles AlI3 1.60 moles AlI3

3 moles I2

• Al is limiting reactant

What is the limiting Reactant with

1.20 grams of Al and 2.40 grams of iodine?A 1.20g Al

B 2.40 g I2

1.20 g Al 1mol Al 2 mol AlI3 .0445 mols AlI3

26.9815 g 2 mol Al 2.40 g I2 1 mol I2 2 mol AlI3 .00630 mols AlI3

253.8089 g 3 mole I2

I2 is limiting Reactant! B

How many grams of excess reactant will remain?

2.40g I2 1 mol I2 2 mols Al 26.9815g

253.8089g 3 mole I2 1 mol Al

.0.170 g of Al reacted Started with 1.20g -0.170g reacted1.03 g in excess

15.00 g aluminum sulfide and 12.00 g of water react until the limiting reagent is used up. Here is the balanced equation for the reaction.

Al2S3 + 6 H2O 2 Al(OH)3 + 3 H2S

a.) What is the maximum mass of H2S which can be formed from these reagents ?

A. 10.21gB. 11.35 g

15.00g Al2S3 1 mol Al2S3 3 mols H2S 34.08 g

150.16167g 1 mol Al2S3 1 mol H2S10.21g H2S

12.00g H2O 1 mol H2O 3 mols H2S 34.08 g 18.008g 6 mol H2O 1 mol H2S11.35g H2S

A. 10.21g

What is the limiting Reactant?• A . H2O• B. Al2S3

• B!

• How much of the excess reactant is used up?

• A, 1.200g• B. 12.00 g• C. 10.84 g

15.00g Al2S3 1 mol Al2S3 6 mols H2O 18.08 g

150.16167g 1 mol Al2S3 1 mol H2010.84 g used up

Percent Yield

Percent Yield =

(Actual Yield / Theoretical Yield) * 100

• Theoretical Yield is the amount that your stoichiometric calculations have predicted.

• Actual Yield is what you actually produced in an experiment.

Example: In the reaction between nitrogen gas and hydrogen gas producing ammonia:

a.) If you start with 14 grams of N2 and 6.0 grams of H2, what will be the limiting reagent and the theoretical yield?

b.) If 10. grams of the product was formed, what would be the percent yield?

Stoichiometry with Solutions

• Just like other problems but use

Moles = Molarity X Liters

to get moles,

or

Liters = Moles/Molarityto find the volume of a solution needed in a

reaction.

How many grams of aluminum will berequired to completely replace copperfrom a 208 mL of a 0.100M solution ofcopper(II) chloride?

2 Al + 3 CuCl2 2 AlCl3 + 3 Cu

Answer = 0.374 g Al

How many milliliters of a 3.40M CuSO4

solution will be needed to react with510.0 g of AgCl in the followingreaction?AgCl + CuSO4 Ag2SO4 + CuCl2

Answer = 523 mL