Calculate the amount of `KClO_(3)` needed to supply sufficient oxygen for burning 112 L of CO gas at N.T.P.

Explanation

Calculation of `O_(2)` gas required to burn 112 L of CO.
`underset(1 mol)(CO) + underset(0.5 mol)(1/2 O_(2)) to CO_(2)`
22.4 L at N.T.P.
22.4 L of CO at N.T.P. require `O_(2) = 0.5` mol
112 L of CO at N.T.P. require `O_(2) = 0.5/(22.4) xx 112`
= 2.5 mol
This is `O_(2)` is to be obtained by heating `KClO_(3)`
`underset(2 "mol")(2KClO_(3)) to 2KCl + underset(3 "mol")(3O_(2))`
3 mol of `O_(2)` is produced from `KClO_(3)= 245 g`
2.5 mol of `O_(2)` is produced from `KClO_(3) = (245 xx 25)/3`
=204.167 g