a)
$2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2$
b)
Bảo toàn khối lượng :
$m_{MnO_2} + m_{hh} = m_{hh\ sau\ pư } + m_{O_2}$
$\Rightarrow m_{O_2} = 3 + 197 - 152 = 48(gam)$
$\Rightarrow n_{O_2} = 1,5(mol)$
Theo PTHH : $n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 1(mol)$
$\%m_{KClO_3} = \dfrac{1.122,5}{197}.100\% = 62,2\%$
$\%m_{KCl} = 100\% - 62,2\% = 37,8\%$