1)
Spu thu đc 76-2= 74g KCl
=> $n_{KCl spu}$= $\frac{74}{74,5}$= 1 mol
Gọi x là mol KCl, y là mol $KClO_3$ ban đầu
=> 74,5x+ 122,5y= 98 (1)
2$KClO_3$ (t*)-> 2KCl+ 3$O_2$
=> $n_{KCl nung}$= y
=> x+y= 1 (2)
(1)(2)=> x= y= 0,5
$m_{KCl}$= 0,5.74,5= 37,25g
$m_{KClO_3}$= 0,5.122,5= 61,25g
2)
2$KNO_3$ (t*)-> 2$KNO_2$ + O2
a,
$n_{O_2}$= 11,2/32= 0,35 mol
=> $n_{KNO_3}$= $\frac{0,35}{80%}$ .101= 44,1875g
b,
$n_{KNO_3}$= $\frac{40,4}{101}$= 0,4 mol
=> $m_{O_2}$= 0,4.32.85%= 10,88g