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When 6.0 mol O2 react with excess H2, 9.0 mol H2O are actually produced, what is the percent yield of the reaction?

A. 25.0%
B. 50.0%
C. 75.0%
D. 100.0%

1 Answer

1 vote

Final answer:

To calculate the percent yield of the reaction, divide the actual yield by the theoretical yield and multiply by 100. In this case, the percent yield is 75.0%.

Step-by-step explanation:

To calculate the percent yield of the reaction, you need to compare the actual yield (9.0 mol H2O) to the theoretical yield.

From the balanced equation 2 H2 + O2 → 2 H2O, you can see that 1 mol of O2 reacts to produce 2 mol of H2O.

Therefore, when 6.0 mol of O2 react, the theoretical yield of H2O would be 12.0 mol. The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.

In this case, the percent yield is (9.0 mol H2O / 12.0 mol H2O) x 100 = 75.0%.

Therefore, the correct answer is C. 75.0%.

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