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)
1.16 g H2 is allowed to react with 10.2 g N2, producing 2.55 g NH3.
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.
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
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.
A)
Molar mass of H2 = 2.016 g/mol
mass(H2)= 1.16 g
use:
number of mol of H2,
n = mass of H2/molar mass of H2
=(1.16 g)/(2.016 g/mol)
= 0.5754 mol
Molar mass of N2 = 28.02 g/mol
mass(N2)= 10.2 g
use:
number of mol of N2,
n = mass of N2/molar mass of N2
=(10.2 g)/(28.02 g/mol)
= 0.364 mol
Balanced chemical equation is:
3 H2 + N2 ---> 2 NH3 +
3 mol of H2 reacts with 1 mol of N2
for 0.5754 mol of H2, 0.1918 mol of N2 is required
But we have 0.364 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)*0.5754
= 0.3836 mol
use:
mass of NH3 = number of mol * molar mass
= 0.3836*17.03
= 6.534 g
Answer: 6.53 g
B)
% yield = actual mass*100/theoretical mass
= 2.55*100/6.534
= 39.0 %
Answer: 39.0 %
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