The percent yield of H2O when 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0 g of H2 is 90.91%. This is calculated using the percent yield equation, which is actual yield divided by theoretical yield multiplied by 100.

## Calculating Theoretical Yield

The theoretical yield is the amount of product produced when all of the limiting reactant is consumed. In this case, the limiting reactant is H2, which has 11.0 g. The equation for the reaction is 2H2 + O2 → 2H2O, and the mole ratio of H2O to H2 is 2:1. This means that for every mole of H2, two moles of H2O are produced.

## Calculating Actual Yield

The actual yield is the amount of product actually produced in the reaction, which in this case is 87.0 g of H2O.

## Calculating Percent Yield

The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. The theoretical yield is 22.0 g (11.0 g of H2 times 2 moles of H2O per mole of H2), and the actual yield is 87.0 g. Therefore, the percent yield is 87.0 g divided by 22.0 g, which is 3.95, multiplied by 100, giving a percent yield of 90.91%.

## Related Questions

• What is the equation for the reaction of 2H2 + O2 → 2H2O?
• What is the mole ratio of H2O to H2?
• What is the theoretical yield of H2O for 11.0 g of H2?
• What is the actual yield of H2O for combining 95.0 g of O2 and 11.0 g of H2?
• What is the percent yield of H2O for 95.0 g of O2 and 11.0 g of H2?
• What is the limiting reactant in the reaction of 2H2 + O2 → 2H2O?
• What is the equation for calculating percent yield?
• What is the unit for percent yield?
• What is the percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 22.0 g of H2?
• What is the percent yield of H2O if 87.0 g of H2O is produced by combining 110.0 g of O2 and 11.0 g of H2?