1.40 g H2 is allowed to react with 10.0 g N2, producing 2.44 g NH3.

Question

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

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

Answer 1

The balanced equation is; N_{2 +} 3H_{2 =} 2NH₃

 

28 (14*2) grams Nitrogen + 6 (3*2) grams Hydrogen gives 34 grams Ammonia

 

1 gram Hydrogen requires 28/6 grams Nitrogen to produce 34/6 grams Ammonia

 

1.4 grams Hydrogen requires 28/6*1.4(6.53) grams Nitrogen to produce 34/6*1.4(7.93) grams Ammonia

 

(A) Hence the theoretical yield is 7.93 grams Ammonia

 

(B) Percentage yield = 2.44/7.93*100% = 30.8%