3H2(g)+N2(g)2NH3(g) The ammonia produced in the Haber-Bosch process has a wide r
ID: 548243 • Letter: 3
Question
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.
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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.
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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 equation3H2(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
Explanation / Answer
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 %
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