When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l) A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%

D) 65.7%

Explanation:

Based on the reaction:

2H2(g)+O2(g)⟶2H2O(l)

2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.

To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.

Theoretical yield:

Moles of 5.58g H₂:

5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂

As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:

2.768 moles H₂O ₓ (18.015g / mol) =

49.86g H₂O is theoretical yield

Percent yield:

Percent yield = Actual yield / Theoretical yield ₓ 100

32.8g H₂O / 49.86g ₓ 100 =

65.7% is percent yield of the reaction

5.6⋅10−^19  j

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