Iron metal reacts with hydrochloric acid as follows: 2Fe(s) + 6HCl(aq) $\rightarrow$ 2FeCl₃(aq) + 3H₂(g) If 35.6 g of iron react with excess HCl, and 98.6 g of FeCl₃ are collected, what is the percent yield of FeCl₃?

Question

Quizplus · Accepted Answer

First, we need to calculate the theoretical yield of FeCl₃ using stoichiometry: 1 mol of Fe reacts with 3 mol of HCl to produce 1 mol of FeCl₃ Molar mass of Fe: 55.85 g/mol Mass of Fe used: 35.6 g Number of moles of Fe used: 35.6 g / 55.85 g/mol = 0.637 mol Theoretical yield of FeCl₃: 0.637 mol Fe x (1 mol FeCl₃/1 mol Fe) x (162.2 g FeCl₃/mol) = 104.0 g FeCl₃ Now we can calculate the percent yield: % yield = (actual yield/theoretical yield) x 100% Actual yield = 98.6 g FeCl₃ (given) % yield = (98.6 g/104.0 g) x 100% = 95.7% Therefore, the answer is C) 95.7%.

Iron Metal Reacts with Hydrochloric Acid as Follows: 2Fe(s) →\rightarrow→ 2FeCl3(aq) + 3H2(g) If 35

Iron Metal Reacts with Hydrochloric Acid as Follows: 2Fe(s) $\rightarrow$ 2FeCl₃(aq) + 3H₂(g) If 35