Question

What is the percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0 g of H2

Answer 1

Answer:- 87.9%

Solution:- The balanced equation for the formation of water from hydrogen and oxygen is:

$2H_2+O_2\rightarrow 2H_2O$

We will calculate the grams of water from the given grams of hydrogen and oxygen and see which one of these two gives least amount of water. This least amount of water will be the theoretical yield.

$95.0gO_2(\frac{1molO_2}{32gO_2})(\frac{2molH_2O}{1molO_2})(\frac{18gH_2O}{1molH_2O})$

= $107gH_2O$

$11.0gH_2(\frac{1molH_2}{2gH_2})(\frac{2molH_2O}{2molH_2})(\frac{18gH_2O}{1molH_2O})$

= $99.0gH_2O$

From above calculations, hydrogen gives least amount of water. So, the theoretical yield of water is 99.0g.

Actual yield of water is given as 87.0g. percent yield is calculated by using the formula:

percent yield = $(\frac{actual}{theoretical})100$

percent yield = [tex](\frac{87.0}{99.0})100

percent yield = 87.9%