Answer:
The percentage yield of water is 66.67%.
Explanation:
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
Mass of oxygen gas = 100 g
Moles of oxygen gas = [tex]\frac{100 g}{32 g/mol}=3.125 mol[/tex]
According to reaction, 1 mole of oxygen gives 2 moles of water, then 3.125 moles of oxygen will give:
[tex]\frac{2}{1}\times 3.125 mol=6.25 mol[/tex]
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 :
[tex]=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{75 g}{112.5 g}\times 100=66.67\%[/tex]
The percentage yield of water is 66.67%.
