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.)

Compute the mass of potassium chlorate (KClO_(3))that should decompose

1.	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.)
Answer» Solution :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 to2KCI + 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