Compute the mass of potassium chlorate `(KClO_(3)) ` that should decompose to produce 8 g of oxygen as per the chemical equation , `2KClO_(3) to 2KCl + 3O_(2)(g)`
(R.A.M : K = 39, Cl = `35.5,O = 16`.)

Explanation

Let us first convert the required volume of oxygen (10.0 L) at S.T.P.
`P_1 = 750 " torr " = 750 mm Hg. " " V_1 = 10.0 L`,
`T_1 = 25 + 273 = 298 K`
and `P_2 = 760 mm Hg " (1 atm)," " " V_2` = ?,
`T_2 = 0^@C = 273 K`
According to the gas equation,
`(P_1 V_1)/T_1 = (P_2 V_2)/T_2`
`:. V_2 = (P_1 V_1 T_2)/(T_1 P_2)=(750 xx 10.0 xx 273)/(298 xx 760)=9.04 L`
Hence, the required volume of `O_2` at S.T.P. = 9.04 L The equation involved is
`2KCIO_3 to 2KCI + 3O_2`
`2xx(39+ 35.5+48) " " 3xx22.4L`
=`245 g " " S.T.P`.
`:. 3 xx 22.4 L " of " O_2 " is produced by the decomposition of " KClO_3 = 245 g`
`:. 9.04 L " of " O_2` will be produced by the
decomposition of `KClO_3 = 245/(3xx22.4)xx 9.04 = 32.9` g
Hence, the required amount of `KClO_3 = 32.9` g