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.

1.73 g H2 is allowed to react with 10.3 g N2, producing 2.34 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. Hints SubmitMy AnswersGive Up Part B What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. Hints

Answer 1

A)

mass(H2)= 1.73 g

Molar mass of H2 = 2.016 g/mol

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.73 g)/(2.016 g/mol)

= 0.8581 mol

mass(N2)= 10.3 g

Molar mass of N2 = 28.02 g/mol

number of mol of N2,

n = mass of N2/molar mass of N2

=(10.3 g)/(28.02 g/mol)

= 0.3676 mol

we have the Balanced chemical equation as:

3 H2 + N2 ---> 2 NH3

3 mol of H2 reacts with 1 mol of N2

for 0.8581 mol of H2, 0.286 mol of N2 is required

But we have 0.3676 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

From balanced chemical reaction, we see that

when 3 mol of H2 reacts, 2 mol of NH3 is formed

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

= (2/3)*0.8581

= 0.5721 mol

mass of NH3 = number of mol * molar mass

= 0.5721*17.03

= 9.745 g

Answer: 9.75 g

B)

% yield = actual mass*100/theoretical mass

= 2.34*100/9.745

= 24.0 %

Answer: 24.0 %