Question

125 grams of HCl are used. Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(g). What is the percent yield if 1.2 moles of hydrogen are actually produced?

Answer 1

Answer:

70%

Explanation:

To find the percent yield, you need to (1) convert grams HCl to moles (via molar mass from periodic table), then (2) convert moles HCl to moles H₂ (via mole-to-mole ratio from equation), then (3) calculate the percent yield H₂ (via percent yield equation).

Zn (s) + 2 HCl (aq) --> ZnCl₂ (aq) + 1 H₂ (g)

Molar Mass (HCl): 1.008 g/mole + 35.45 g/mole
Molar Mass (HCl) = 36.458 g/mole

125 g HCl         1 mole HCl         1 mole H₂
---------------  x  ------------------  x  ------------------  = 1.7 moles H₂
                         36.458 g           2 moles HCl

(actual yield / theoretical yield) x 100% = percent yield

theoretical/calculated yield = 1.7 moles H₂
actual yield = 1.2 moles H₂

(1.2 moles H₂ / 1.7 moles H₂) x 100% = 71%

Therefore, the best percent yield of hydrogen produced is 70%.