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