Question

Baking soda (NaHCO₃) can be made in large quantities by the following reaction

NaCl(aq)+NH₃(aq)+CO₂(aq)+H₂(l) → NaHCO₃(s)+NH₄Cl(aq)

If 9.87 g of nacl reacts with excesses of the other reactants and 3.57 g of NaHCO₃ is isolated, what is the percent yield of the reaction?

Answer 1

The percent yield of NaHCO3 is calculated by dividing the actual yield (3.57 g) by the theoretical yield (14.198 g), then multiplying by 100%, which results in a percent yield of approximately 25.15%.

Step-by-step explanation:

To determine the percent yield of NaHCO₃, we must first calculate the theoretical yield based on the stoichiometry of the given reaction. Since the molar mass of NaCl is approximately 58.44 g/mol, we can find out how many moles of NaCl reacted by dividing the mass of reactant (9.87 g) by its molar mass:

9.87 g NaCl / 58.44 g/mol = 0.169 mol NaCl

According to the balanced chemical equation, 1 mole of NaCl yields 1 mole of NaHCO₃. Therefore, the theoretical yield of NaHCO₃ would be 0.169 mol. The molar mass of NaHCO₃ is approximately 84.01 g/mol, so the theoretical yield in grams is:

0.169 mol × 84.01 g/mol = 14.198 g of NaHCO₃ (theoretical yield)

To find the percent yield, we use the actual yield (3.57 g) and the theoretical yield (14.198 g) in the following formula:

Percent Yield = (Actual Yield / Theoretical Yield) × 100%

Percent Yield = (3.57 g / 14.198 g) × 100% ≈ 25.15%

Therefore, the percent yield of the reaction that produced NaHCO₃ is approximately 25.15%.