Question

Consider the chemical equation for the production of water: 2 H2+O2→2 H2O. If 100 grams of oxygen gas are used, what would the percent yield be if 75 g of H2O was produced? Show your work.

Answer 1

The percentage yield of water is 66.67%.

Step-by-step explanation:

`2H_2+O_2\rightarrow 2H_2O`

Mass of oxygen gas = 100 g

Moles of oxygen gas =
`(100 g)/(32 g/mol)=3.125 mol`

According to reaction, 1 mole of oxygen gives 2 moles of water, then 3.125 moles of oxygen will give:

`(2)/(1)* 3.125 mol=6.25 mol`

Mass of 6.25 moles of water :

6.25 mol × 18 g/mol = 112.5 g

Theoretical yield of water = 112.5 g

Experimental yield of water = 75 g

Percentage yield :

`=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`=(75 g)/(112.5 g)* 100=66.67\%`

The percentage yield of water is 66.67%.