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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)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.1.29g H2 is allowed to react with 9.55g N2, producing 1.49g NH3.

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

User Adam Musch
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1 Answer

1 vote

Answer: a. 7.31 g

b. 20.4 %

Step-by-step explanation:

To calculate the moles :


\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}


\text{Moles of} H_2=(1.29g)/(2g/mol)=0.645moles


\text{Moles of} N_2=(9.55)/(28g/mol)=0.341moles


3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

According to stoichiometry :

3 moles of
H_2 require = 1 mole of
N_2

Thus 0.645 moles of
H_2 will require=
(1)/(3)* 0.645=0.215moles of
N_2

Thus
H_2 is the limiting reagent as it limits the formation of product and
N_2 is the excess reagent.

As 3 moles of
H_2 give = 2 moles of
NH_3

Thus 0.645 moles of
H_2 give =
(2)/(3)* 0.645=0.430moles of
NH_3

Mass of
NH_3=moles* {\text {Molar mass}}=0.430moles* 17g/mol=7.31g

Thus theoretical yield for this reaction under the given conditions is 7.31 g.

b)
{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}* 100\%


{\text {percentage yield}}=(1.49g)/(7.31g)* 100\%=20.4\%

The percent yield for this reaction under the given conditions is 20.4 %

User Althaf Hameez
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