Consider the electrolysis of a brine solution: 2NaCl(aq) + H₂O $\rightarrow$ Cl₂(g) + H₂(g) + 2NaOH(aq)
A 10.0-g sample of sodium chloride was placed in 10.0 g of water. If 3.85 g of Cl₂ was obtained, what was the percent yield of Cl₂?

Question

Consider the electrolysis of a brine solution: 2NaCl(aq) + H₂O

\rightarrow

Cl₂(g) + H₂(g) + 2NaOH(aq)
A 10.0-g sample of sodium chloride was placed in 10.0 g of water. If 3.85 g of Cl₂ was obtained, what was the percent yield of Cl₂?

Quizplus · Accepted Answer

The theoretical yield of Cl₂ can be calculated from the moles of NaCl, which is the limiting reactant. The molar mass of NaCl is 58.44 g/mol, so 10.0 g of NaCl is 0.171 mol. According to the balanced equation, 1 mol of NaCl produces 0.5 mol of Cl₂. Therefore, 0.171 mol of NaCl produces 0.0855 mol of Cl₂. The molar mass of Cl₂ is 70.90 g/mol, so the theoretical yield is 0.0855 mol × 70.90 g/mol = 6.06 g. The percent yield is (3.85 g / 6.06 g) × 100% = 63.5%.

Consider the Electrolysis of a Brine Solution: 2NaCl(aq) + H2O →\rightarrow→

Consider the Electrolysis of a Brine Solution: 2NaCl(aq) + H₂O $\rightarrow$