Question

What is the percent yield for this reaction under the given conditions? Express answers to three significant figures and include the appropriate units.

1.74g H2 is allowed to react with 10.2g N2 producing 1.90g NH3

Answer 1

balanced chemical reaction

3H2(g) + N2(g) → 2NH3(g)

find the mole of H2 = 1.74 g / 2.016 g/mol= 0.863 mole of H2

and the mole of N2 = 10.2g / 28.0 g/mol = 0.364 moles of N2

find the limiting reagent

3 mole of H2 react with 1 mole of N2 produce 2 mole of NH3

0.863 moles of H2 react with (0.863 / 3) = 0.2876 moles of N2

but N2 is in exece amount 0.364 moles

so H2 is limiting reagent

0.863 moles of H2 produce (0.863 x 2/3) = 0.57533 moles of NH3

H2 is the limiting reactant, and the theoretical yield of NH3 = 0.57533 mol NH3

mass of produce NH3 = (0.57533 mol x 17.031g/mol) = 9.78 g NH3

% yield = (actual / theoretical) X 100
% yield = (1.90 g NH3 / 9.78 g NH3) X 100 = 19.426 %