Question

What is the percent yield of the reaction when 16.0 g of methane reacts with excess oxygen to give 8.00 g of carbon dioxide? The balanced equation is CH4(g) + 2O2 (g) -> CO2(g) +2H2O(g)

Answer 1

The balanced equation for the reaction is given.

CH₄ (g) + 2 O₂ (g) -------> CO₂ (g) + 2 H₂O (g)

As per the balanced stoichiometric equation,

1 mole CH₄ = 1 mole CO₂.

Molar mass of CH₄ = (1*12.01 + 4*1.008) g/mol = 16.042 g/mol.

Moles of CH₄ = (16.0 g)/(16.042 g/mol) = 0.9974 mole.

Therefore, moles of CO₂ generated = moles of CH₄ reacted = 0.9974 mole.

Molar mass of CO₂ = (1*12.01 + 2*15.9994) g/mol = 44.0088 g/mol.

Therefore, mass of CO₂ theoretically expected = (0.9974 mole)*(44.0088 g/mol) = 43.8944 g.

However, the yield of CO₂ was 8.00 g.

Therefore, the percent yield of CO₂ = (actual yield)/(theoretical yield)*100 = (8.00 g)/(43.8944 g)*100% = 18.2255% ≈ 18.22% (ans).