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.
1.16 g H2 is allowed to react with 10.4 g N2, producing 2.93 g NH.
What is the percent yield for this reaction under the given conditions?
To calculate the percent yield, we first need to determine the theoretical yield of ammonia based on the given amounts of reactants.
1. Convert the mass of hydrogen gas (H2) and nitrogen gas (N2) to moles:
1.16 g H2 * (1 mol H2 / 2.02 g H2) = 0.574 moles H2
10.4 g N2 * (1 mol N2 / 28.02 g N2) = 0.371 moles N2
2. Use the stoichiometry of the balanced chemical equation to determine the limiting reactant and the theoretical yield of ammonia:
From the equation: 3 moles of H2 react with 1 mole of N2 to produce 2 moles of NH3
0.371 moles of N2 * (3 moles H2 / 1 mole N2) = 1.113 moles H2 needed
Since there is only 0.574 moles of H2, it is the limiting reactant and will determine the theoretical yield of ammonia:
0.574 moles H2 * (2 moles NH3 / 3 moles H2) = 0.383 moles NH3
3. Convert the theoretical yield of ammonia from moles to grams:
0.383 moles NH3 * (17.03 g NH3 / 1 mol NH3) = 6.51 g NH3
4. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100%
Percent yield = (2.93 g NH3 / 6.51 g NH3) * 100% = 45.1%
Therefore, the percent yield for this reaction under the given conditions is 45.1%.