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

90.9% is the percent yield of the reaction.

Step-by-step explanation:

Mass of hydrogen gas = 24.2 g

Moles of hydrogen =
`(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:

`(2)/(2)* 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}}* 100`

`=(198 g)/(217.8 g)* 100=90.9\%`

90.9% is the percent yield of the reaction.