`a)`

`2KClO_3` $\xrightarrow{t^o}$ `2KCl+3O_2`

`n_{KCl}={14,9}/{74,5}=0,2(mol)`

`n_{KClO_3}=n_{KCl}=0,2(mol)`

`m_{KClO_3}=0,2.122,5=24,5(g)`

`b)`

`n_{O_2}=3/2n_{KCl}=0,3(mol)`

`V_{O_2}=0,3.22,4=6,72(L)`

`c)`

`Hh:Al:x(mol);Fe:y(mol)`

`->27x=56.2y(1)`

`4Al+3O_2` $\xrightarrow{t^o}$ `2Al_2O_3`

`3Fe+2O_2` $\xrightarrow{t^o}$ `Fe_3O_4`

`->n_{O_2}=0,75x+2/{3}y=0,3(2)`

`(1)(2)->x={28}/{85};y={27}/{340}`

`->m_{hh}={28}/{85}.27+{27}/{340}.56=13,34(g)`

`n_{KClO_3}=(36,75)/(122,5)=0,3(mol)`

`n_{KCl}=(14,9)/(74,5)=0,2(mol)`

`a,` PTHH:

$ 2KClO_3\xrightarrow{t^o,xt} 2KCl+3O_2$

`=>n_{KClO_3(\text{phản ứng})}=  n_{KCl}= 0,2(mol)`

`=>m_{KClO_3(\text{phản ứng})}=0,2.122,5=24,5(g)`

`b,`

`n_{O_2}= 3/2 n_{KCl}= 3/2 . 0,2=0,3(mol)`

`=>V_{O_2}=0,3.22,4=6,72(l)`

`c,`

Gọi `n_{Fe}=a(mol)`

`=>m_{Fe}=56a(g)=>m_{Al}=56a.2=112a(g)`

`=>n_{Al}=(112a)/(27)(mol)`

PTHH:

$\rm 3Fe+2O_2\xrightarrow{t^o}Fe_3O_4 \\ 4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

`=>n_{O_2}= 2/3 n_{Fe} + 3/4 n_{Al}= 2/3 . a + 3/4 . (112a)/(27)=(34)/9 a =0,3`

`=>a=(27)/(340)(mol)`

`=>n_{Fe}=(27)/(340)(mol);n_{Al}=(27)/(85)(mol)`

`=>m_{hh}=(27)/(340).56+(27)/(85).27≈13,02(g)`