Final answer:
To determine the percent yield of the reaction, calculate the theoretical yield by stoichiometry and then compare it to the actual yield. The percent yield is approximately 79.5%.
Step-by-step explanation:
To determine the percent yield of the reaction, we need to first calculate the theoretical yield and then compare it to the actual yield. The balanced equation for the reaction is:
2NaCl + NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
The molar mass of NaCl is 58.44 g/mol and the molar mass of NaHCO₃ is 84.01 g/mol. From the given masses, we can calculate the number of moles of NaCl and NaHCO₃:
Moles of NaCl = 9.29 g / 58.44 g/mol = 0.159 mol
Moles of NaHCO₃ = 4.01 g / 84.01 g/mol = 0.048 mol
The stoichiometry of the balanced equation tells us that 2 moles of NaCl react with 1 mole of NaHCO₃. Therefore, the limiting reactant is NaHCO₃ because it would require 0.096 moles of NaHCO₃ to react with the 0.159 moles of NaCl present. Since only 0.048 moles of NaHCO₃ is available, it is the limiting reactant.
Now, to calculate the theoretical yield of Na₂CO₃, we can use the mole ratio:
Theoretical yield of Na₂CO₃ = 0.048 mol NaHCO₃ × (1 mol Na₂CO₃ / 1 mol NaHCO₃) × (105.99 g Na₂CO₃ / 1 mol Na₂CO₃) = 5.039 g
The actual yield given is 4.01 g NaHCO₃, so the percent yield can be calculated as:
Percent yield = (Actual yield / Theoretical yield) × 100 = (4.01 g / 5.039 g) × 100 = 79.5%
Therefore, the correct answer is not listed, but the percent yield is approximately 79.5%.