Calculate the amount of potassium chlorate `(KClO_(3))` which must be heated to produce `2.4` litres of oxygen at `740mm` `Hg` and at `25^(@)C`.

Explanation

Step I. Calculation of volume of oxygen evolved at `N.T.P.`
`V_(1) = 2.4 L` , `V_(2) = ?`
`P_(1) = 74 mm = (740)/(760) "bar"` , `P_(2) = 1.013` bar
`T_(1) = 25+273 = 298 K` , `T_(2) = 273 K`
According to gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))`
Substituting the values, `V_(2) = ((740)/(460) ("bar") xx 2.4 (L) xx 273 (K))/(298 (K) xx 1.013 ("bar")) = 2.113 L`
Step II. Calculation of amount of `KClO_(3)`.
`underset(=245 g)underset(2(39+35.5+48))(2KClO_(3))overset("Heat")rarr2KCl+underset(=67.2 L)underset(3xx22.4)(3O_(2))`
`67.2 L` of `O_(2)` is envloved from `KClO_(3) = 245 g`
`1.0 L` of `O_(2)` is evolved from `KClO_(3) = (245)/(67.2) g`
`2.113 L` of `O_(2)` evolved from `KClO_(3) = (245)/(67.2) xx 2.113 =7.7g`