Question

During a laboratory experiment, a 3.81-gram sample of NaHCO₃ was thermally decomposed. In this experiment, carbon dioxide and water vapors escape and are combined to form carbonic acid. After decomposition, the sample weighed 2.86 grams. Calculate the percentage yield of carbonic acid for the reaction. Describe the calculation process in detail.

NaHCO₃ → Na₂CO₃ + H₂CO₃

Answer 1

molecular weight of NaHCO₃ = 84 g/mol (23 + 1 + 12 + 3 × 16)

molecular weight of H₂CO₃ = 62 g/mol (1 × 2 + 12 + 3 × 16)

3.81 gm NaHCO₃ = 3.81 g / 84 g.mol^-1= 0.045 mol

2.86 gm NaHCO₃ = 2.86 g / 84 g.mol^-1= 0.034 mol

Moles of NaHCO₃ reacted = 0.011 mol

Balanced equation: NaHCO₃ = 0.5 Na₂CO₃ + 0.5 H₂CO₃​​​​​​

1 mol of NaHCO₃ produces 0.5 mol H₂CO₃​​​​​​

Moles of H₂CO₃ produced = 0.5 × 0.011 = 0.0055 mol = 0.0055 mol × 62 g/mol = 0.341 g

Actual yield = 0.341 g

If whole of NaHCO₃ have reacted then H₂CO₃ produced = 0.045 mol × 0.5 × 62 g/mol = 1.395 g

Theoretical yield = 1.395 g

% yield = actual yield / theoretical yield × 100 = 0.341 / 1.395 × 100 = 24.44%