10.The reaction of 10.0 g H2 with 10.0 g O2 yields 7.81 g H2O. What is the percent yield of this reaction?

Answer 1

First, make a balanced equation:

2H2 + 02 → 2H2O

Now find the theoretical yield (use O2 because it's the limiting reagent):

10.0 g O2 * (1 mol O2/31.9988 g O2) = 0.313 mol O2

0.313 mol O2 * (2 mol H2O/1 mol O2) = 0.626 mol H2O

0.626 mol H2O * (18.009 g/1 mol H2O) = 11.3 g H2O (this is the theoretical yield)

Percent yield = (actual yield/theoretical yield) * 100

Percent yield = (7.81/11.3) * 100

Percent yield = 69.1%


* Please vote my answer as the best since I went into greater detail. Thanks.

Answer 2

10.0 g H2 = 4.96 mol H2
10.0 g O2 = 0.313 mol O2
2 H2 + O2 --> 2 H2O
Oxygen is the limiting reagent.
.313 mol O2 = .626 mol H2O = 11.3 g H2O
7.81/11.3 = 69.1%