Question

1.33 g H2 is allowed to react with 9.73 g N2, producing 1.40 g NH3.

Part A

What is the theoretical yield in grams for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Part B

What is the percent yield for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer for part A

Given data,

1.33g of H2 = 0.665 moles

9.73g of N2 = 0.347 moles

Produces 1.40g of NH3 = 0.082 moles

In this reaction N2 is limiting reactant as it has least no of reactant moles so it will consumed and forms equal no of product moles.

Theoretically 0.347 moles of NH3 should form.

Theoretical yield = 0.347 moles of NH3 = 5.899g of NH3

Answer for part B

Percent yield = (practical yield/theoretical yield)×100

= (1.40/5.899)×100

= 23.73% is the percent yield of the given reaction.