Question

5.0 kg of hydrogen is reacted with 31 kg of oxygen. The experiment yields 28 kg of water. Calculate the percent yield of the reaction

Answer 1

2H₂ + O₂ → 2H₂O

M(H₂) = 2 g/mol

n(H₂) $= \frac{5000}{2} = 2500 \;mol$

M(O₂) = 32 g/mol

n(O₂) $= \frac{31000}{32} = 968.75 \;mol$

According to the reaction for one mole of O₂ we need two moles of H₂. So, O₂ is limiting reactant.

n(H₂O) = 2n(O₂) $= 2 \times 968.75 = 1937.5 \;mol$

M(H₂O) = 18 g/mol

m(H₂O)(theoretical) $= 1937.5 \times 18 = 34875 \;g = 34.875 \;kg$

Proportion:

37.875 kg – 100 %

28 kg – x

$x = \frac{28 \times 100}{37.875} = 73.927 \; \%$

Answer: 74 %