Chemistry Lab questions

Question

Ammonia NH3, can be synthesized by the following reaction: 2NO + 5H2 → 2NH3 +2 H2O

1.) If you start with 90.0 grams of NO and an unlimited supply of H2, what is the theoretical yield of ammonia? How much water is produced?

2.) What is the percent yield if 25 grams of ammonia are produced?

3.) If 5 grams of product are produced out of a possible 15 grams, what is the percent yield of that product?

Answer 1

REACTION:

2NO + 5H₂ → 2NH₃ +2 H₂O

 

1.) If you start with 90.0 grams of NO and an unlimited (excess) supply of H₂, what is the theoretical yield of ammonia?

 

Limiting reagent = NO

Molar mass of NO = 1(14) + 1(16) = 30 g/mol

 

(90 g NO) / (30 g/mol NO) = 3 mol NO

 

Stoichiometry: 2 NO produces 2 NH₃  (1:1 ratio)

Therefore;

3 mol NH₃ should be produced

 

Molar mass of NH₃ = 1(14) + 3(1) = 17 g/mol

 

3 mol NH₃ * 17 g/mol = 51 g NH₃ (Theoretical yield)

 

 

How much water is produced?

 

Stoichiometry: 2 NO produces 2 H₂O (1:1 ratio)

Therefore;

3 mol H₂O will be produced 

 

Molar mass H₂O = 1(16) + 2(1) = 18 g/mol

3 mol H₂O * 18 g/mol H₂O = 54 g H₂O

 

2.) What is the percent yield if 25 grams of ammonia are produced?

 

Percent yield = 100 × actual yield/theoretical yield

 

Percent yield NH₃ = 100 × 25 g / 51 g

percent yield NH₃ = 49%

 

3.) If 5 grams of product are produced out of a possible 15 grams, what is the percent yield of that product?

 

percent yield product = 100 × 5 g / 15 g

 

percent yield product = 33%