A reaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is charged with 5.79 kg of h2 and excess n2. a total of 26.8 kg of nh3 are produced. what is the percent yield of the reaction?

82.08 %

Explanation:

The percent yield of the reaction = [ (actual yield) / (calculated yield) ] x 100.

Actual yield = 26.80 kg.

To get the calculated yield: The balanced equation of reacting N2 with H2 to produce NH3 is:

N₂ + 3H₂ → 2NH₃

It is clear that 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃. N₂ is present in excess and H₂ is the limiting reactant. We need to convert the mass of H₂ added (5.79 kg) to moles using the relation:

n = mass / molar mass = (5790 g) / (2.01 g/mol) = 2880.6 mol.

We can get the no. of moles of NH₃ produced.

Using cross multiplication:

3.0 mole of H₂ produce → 2.0 moles off NH₃, from the stichiometry.

2880.6 mol of H₂ produces →? moles of NH₃.

The no. of moles of NH₃ produced = (2880.6 mol) (2.0 mol) / (3.0 mol) = 1920.4 mol. We can know get the calculated yield of NH₃ = no. of moles x molar mass = (1920.4 mol) (17.00 g/mol) = 32646.76 g ≅ 32.65 kg.

∴ The percent yield of the reaction = [ (actual yield) / (calculated yield) ] x 100 = [ (26.8 kg) / (32.65 kg) ] x 100 = 82.08 %.