Question

The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation

3H2(g)+N2(g)→2NH3(g)

The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation.

1.60 g H2 is allowed to react with 10.3 g N2, producing 2.24 g NH3.

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

percentage yield

H2 = 2g/mol
1.60g/2g/mol = 0.8 mol H2
N2 = 28 g/mol
10.3g / 28 g/mol = 0.367mol N2

0.367mol N2 required 1.101 mol of H2
=> H2 is the limiting reagent as to react all of the N2 you would require 1.101 mol of H2



NH3 =17g/mol

for every 0.8 mol H2 you only require 0.266 of N2 react so the limiting reagent is Hydrogen

0.367*3 = 1.101 H2 needed for 0.367 moles of N2

2/3(0.8) *17 = 9.06
possible weight of NH3 =

0.8moles H2 x (2 moles NH3 / 3 mole H2) x (17.03 g NH3 / 1 mole NH3)
=9.06 grams
yield =(actual ammonia produced /theoritical mass )*100 = (2.24/9.06)*100/1=24.72 %