Answer:
The percent yield of oxygen in this chemical reaction is 73.71 %.
Explanation:
Experimental yield of oxygen gas = 115.0 g
Theoretical yield:
[tex]2KClO_3rightarrow 2KCI + 3O_2[/tex]
Mass of potassium chlorate = 400.0 g
Moles of potassium chlorate = [tex]\frac{400 g}{122.5 g/mol}=3.2653 mol[/tex]
According to reaction, 2 moles of potassium chlorate gives 3 moles of oxygen gas.
Then 3.2653 mol of potassium chlorate will give:
[tex]\frac{3}{2}\times 3.2653 mol=4.89795 mol[/tex]
Mass of oxygen gas :
[tex]32 g/mol\times 4.89795 mol=156.73 g[/tex]
Percentage yield:
[tex]\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]\%=\frac{115.0 g}{156.73 g}\times 100=73.71\%[/tex]
The percent yield of oxygen in this chemical reaction is 73.71 %.
