Question

The Haber-Bosch process is a very important industrial process. In the Haber 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 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.

1.57 g H2 is allowed to react with 9.87 g N2, producing 1.69 g Nh3.

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

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

Express both answers to three significant figures and include the appropriate units.

Answer 1

Sol.

Reaction :

N2(g) + 3H2(g) -----> 2NH3(g)

As Mass of N2 = 9.87 g

Molar Mass of N2 = 28.0134 g/mol

So , Moles of N2 = 9.87 g / 28.0134 g/mol

= 0.3523 mol

Now , Mass of H2 = 1.57 g

Molar Mass of H2 = 2.0158 g/mol

So , Moles of H2 = 1.57 g / 2.0158 g/mol

= 0.7788 mol  

From reaction , 1 mole of N2 combines with 3 moles of H2

So , 0.3523 moles of N2 combines with

= 0.3523 × 3 = 1.0569 moles of H2

But we have 0.7788 moles of H2  

Therefore , limiting reactant is H2 and excess reactant is N2

Now , 3 moles of limiting reactant , H2 gives 2 moles of NH3

So , 0.7788 moles of H2 gives  

= 0.7788 × 2 / 3 = 0.5192  moles of NH3  

As Molar Mass of NH3 = 17.031 g/mol

So , Mass of NH3

= 0.5192 mol × 17.031 g/mol

= 8.84 g

Theoretical yield =   8.84 g  

Experimental yield = 1.69 g

Therefore , Percentage yield

= ( Experimental yield / Theoretical yield ) × 100

= ( 1.69 g / 8.84 g ) × 100

=   19.1 %