Iron metal reacts with hydrochloric acid as follows: 2Fe( s ) + 6HCl( aq ) ? 2Fe

Question

Iron metal reacts with hydrochloric acid as follows: 2Fe(s) + 6HCl(aq) ? 2FeCl3(aq) + 3H2(g) If 22.4 g of iron react with excess HCl, and 59.4 g of FeCl3 are collected, what is the percent yield of the reaction?
A. 73.0% B. 109% C. 65.0% D. not enough information given E. 91.4% Iron metal reacts with hydrochloric acid as follows: 2Fe(s) + 6HCl(aq) ? 2FeCl3(aq) + 3H2(g) If 22.4 g of iron react with excess HCl, and 59.4 g of FeCl3 are collected, what is the percent yield of the reaction? Iron metal reacts with hydrochloric acid as follows: 2Fe(s) + 6HCl(aq) ? 2FeCl3(aq) + 3H2(g) If 22.4 g of iron react with excess HCl, and 59.4 g of FeCl3 are collected, what is the percent yield of the reaction? 2Fe(s) + 6HCl(aq) ? 2FeCl3(aq) + 3H2(g) If 22.4 g of iron react with excess HCl, and 59.4 g of FeCl3 are collected, what is the percent yield of the reaction? 73.0% 109% 65.0% not enough information given 91.4% A. 73.0% B. 109% C. 65.0% D. not enough information given E. 91.4%

Explanation / Answer

The answer is: E. 91.4%


2 Fe(s) + 6 HCl(aq) => 2 FeCl3(aq) + 3 H2(g)

Moles of FeCl3 = Moles of Fe = mass/molar mass of Fe

= 22.4/55.85 = 0.401 mol


Theoretical yield = moles x molar mass of FeCl3

= 0.401 x 162.2 = 65.0 g


Percent yield = actual yield/theoretical yield x 100%

= 59.4/65.0 x 100%

= 91.4%

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.