Justin added concentrated HCl solution in excess to 45.8 grams of magnesium metal and collected the hydrogen gas that was produced by the reaction.

Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)

If Justin collected 2.01 grams of H2, what was the percent yield of H2?

A 4.39%

B 46.9%

C 52.9%

D 93.8%

Question

Chemistry

Erik Pearson

5 September, 11:44

Justin added concentrated HCl solution in excess to 45.8 grams of magnesium metal and collected the hydrogen gas that was produced by the reaction.

Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)

If Justin collected 2.01 grams of H2, what was the percent yield of H2?

A 4.39%

B 46.9%

C 52.9%

D 93.8%

+5

Answers (2)

Know the Answer?

Daniel · Answer 1

I have the same question

Armando Sweeney · Answer 2

52.9 %.

Explanation:

Mg + 2HCl - --> MgCl2 + H2

Using the relative atomic masses of magnesium and hydrogen:

Theoretically 24.3 g of Mg will produce 2.016 g of H2

so 45.8 g will produce (2.016 * 45.8) / 24.3 = 3.80 g H2.

So the % yield = 2.01 * 100 / 3.80

= 52.9%.

Justin added concentrated HCl solution in excess to 45.8 grams of magnesium metal and collected the hydrogen gas that was produced by the reaction.Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)If Justin collected 2.01 grams of H2, what was the percent yield of H2?A 4.39%B 46.9%C 52.9%D 93.8%