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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?

125 Grams Of HCl Are Used Zns 2HClaq ZnCl2aq H2g What Is The Percent Yield If 12 Moles Of Hydrogen Are Actually Produced class=

Sagot :

CrystalQueenidn

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%.

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