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 `KClO_(3)`
B. `122*5 g` of `KClO_(3)`
C. `245 g` of `KClO_(3)`
D. `98*0` g of `KClO_(3)`

Answer

A

Explanation

`underset(2xx122*5g)(2KClO_(3))to2KCl+underset(3xx32g)(3O_(3))`
`3xx32 g` of `O_(2)` is produced from `KClO_(3) = 245g`
48 g of `O_(2)` is produced from `KClO_(3) = 122*5g`
`:. 80% KClO_(3)` neededc `=(122.5gxx100)/(80) = 153.12g`