Question

If 24.2 g of hydrogen react with excess oxygen and 198 g of water are produced what is the percent yield?

Answer 1

Answer:

90.9% is the percent yield of the reaction.

Explanation:

Mass of hydrogen gas = 24.2 g

Moles of hydrogen = $\frac{24.2 g}{2g/mol}=12.1 mol$

$2H_2+O_2\rightarrow 2H_2O$

According to reaction, 2 moles of hydrogen gas gives 2 moles of water , then 12.1  moles of hydrogen will give:

$\frac{2}{2}\times 12.1mol=12.1mol$  water

Mass of 12.1 moles of water

= 12.1 mol × 18 g/mol = 217.8 g

Theoretical yield of water = 217.8 g

Experimental yield of water = 198 g

The percentage yield of reaction:

$=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{198 g}{217.8 g}\times 100=90.9\%$

90.9% is the percent yield of the reaction.