Question

During a laboratory experiment, a 2.36-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 1.57 grams. Calculate the percentage yield of carbonic acid for the reaction. Describe the calculation process in detail.
NaHCO3 → Na2CO3 + H2CO3

Answer 1

Answer:

• 90.7 %

Explanation:

1) Chemical equation (given)

• 2NaHCO₃ → Na₂CO₃ + H₂CO₃

2) Theoretical yield

a) Convert mass of NaHCO₃ to moles:

• n = mass in grams / molar mass
• molar mass = 84.007 g/mol
• n = 2.36 g / 84.007 g/mol = 0.02809 mol

b) Mole ratio:

• 2 mol NaHCO₃ : 1 mol H₂CO₃

c) Proportionality:

• 2 mol NaHCO₃ / mol H₂CO₃ = 0.02809 mol NaHCO₃ / x

       ⇒ x = 0.2809 / 2 mol H₂CO₃ = 0.01405 mol H₂CO₃

3) Actual yield

a) Mass balance: 2.36 g - 1.57 g = 0.79 g

b) Convert 0.79 g of carbonic acid to number of moles:

• n = mass in grams / molar mass

• molar mass = 62.03 g/mol

• n = 0.79 g / 62.03 g/mol = 0.01274 mol

4) Percentage yield, y (%)

• y (%) = actual yield / theoretical yield × 100

• y (%) = 0.1274 mol / 0.1405 mol × 100 = 90.68%

The answer must show 3 significant figures, so y(%) = 90.7%.