1g of KClO3 was heated under such conditions such that a part of it decomposes as 2KClO3 gives 2KCl and 3O2,while the remaining part decomposes as 4KClO3 gives 3KClO4 and KCl,if net oxygen obtained is 146.8ml at stp, calculate the mass of KClO4 in the residue.
(0.1458 L O2) / (22.414 L/mol) x (2/3) x (122.5498 g KClO3/mol) =
0.5314 g KClO3 participated in the first reaction
1 g - 0.5314 g = 0.4686 g KClO3 decomposed according to the second reaction
4 KClO3 ? 3 KClO4 + KCl
(0.4686 g KClO3) / (122.5498 g KClO3/mol) x (3/4) x (138.5492 g KClO4/mol) = 0.3973 g KClO4 formed in the second reaction