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:
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%.