Question

21.

In the Haber

-

Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia

according to the following equation

3 H

2

(g) + N

2

(g) → 2 NH

3

(g)

The production of ammonia is difficult and often results in lower yields than those predicted

from the chemical equation. For instance, if 1.36 g of hydrogen gas are allowed to react with

9.97 g of nitrogen gas, 2.24 g of ammonia are produced. What is the

theoretical yield

of the

reaction?

- Not sure why this copied over weird but could i get a detailed walk through for this problem.

Answer 1

1)

Molar mass of H2 = 2.016 g/mol

mass(H2)= 1.36 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.36 g)/(2.016 g/mol)

= 0.6746 mol

Molar mass of N2 = 28.02 g/mol

mass(N2)= 9.97 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(9.97 g)/(28.02 g/mol)

= 0.3558 mol

Balanced chemical equation is:

3 H2 + N2 ---> 2 NH3 +

3 mol of H2 reacts with 1 mol of N2

for 0.6746 mol of H2, 0.2249 mol of N2 is required

But we have 0.3558 mol of N2

so, H2 is limiting reagent

we will use H2 in further calculation

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

According to balanced equation

mol of NH3 formed = (2/3)* moles of H2

= (2/3)*0.6746

= 0.4497 mol

use:

mass of NH3 = number of mol * molar mass

= 0.4497*17.03

= 7.661 g

Answer: Theoretical yield = 7.66 g

2)

% yield = actual mass*100/theoretical mass

= 2.24*100/7.661

= 29.24%

Answer: percent yield = 29.2 %