If potassium chlorate is 80% pure, then 48 g of oxygen would be produced from (atomic mass of K = 39)

Options

A. 153.12 g of `KCIO_(3)`
B. 122.5 g of `KCIO_(3)`
C. 245 g of `KCIO_(3)`
D. 98 g of `KCIO_(3)`

Answer

A

Explanation

`underset(underset(245)(2 xx 122.5 g))(2KClO_(3)) overset("heat")to 2KCl+underset(underset(96)(3 xx 32 g))(3O_(2))`
48 g of oxygen will be produced from 122.5 g of `KClO_(3)`
`:.` Amount of 80% `KClO_(3)` needed
`=(100)/(80) xx 122.5=153.12 g`