a, $2KClO_3 \buildrel{{t^o}}\over\longrightarrow 2KCl+ 3O_2$
b,
$V_{O_2}= 2.6.280= 3360ml= 3,36l$
=> $n_{O_2}= 0,15 mol$
=> $n_{KClO_3}= 0,1 mol$
=> $m_{KClO_3}= 122,5.0,1= 12,25g$
c,
Theo lí thuyết, $0,1 mol KClO_3$ tạo $0,1 mol KCl$
Hao hụt 10% nên chỉ thu đc $0,1.90\%= 0,09 mol KCl$
=> $m_{KCl}= 74,5.0,09= 6,705g$