Đáp án:

$\rm a)\ V_{O_2}=5,04(l)$
$\rm b)$ `%`$\rm m_{KClO_3(du)}$`~~9,88%\ ; %`$\rm m_{KCl}$`=90,12%`

Giải thích các bước giải:

$\rm a)$

$\rm m_{\text{giảm}}=m_{O_2}=7,2(g)$

$\rm n_{O_2}=\dfrac{7,2}{32}=0,225(mol)$

$\rm V_{O_2}=0,225.22,4=5,04(l)$

$\rm b)$

            $\rm 2KClO_3$ $\xrightarrow{xt,t^o}$ $\rm 2KCl+3O_2$
$\rm (mol)$   ` 0,15`                `0,15`   $(0,225)$
$\rm m_{KClO_3(pu)}=0,15.122,5=18,375(g)$

`%`$\rm m_{KClO_3(du)}=\dfrac{19,6-18,375}{(19,6-18,375)+(0,15.74,5)}.100$`%~~9,88%`
`%`$\rm m_{KCl}=$`100%-9,88%=90,12%`

`#Ly`

$a)$

Ta có: Khối lượng giảm $=$ Khối lượng khí thoát ra

PTHH: $2KClO_3$ $\xrightarrow[]{t^o}$ $2KCl$ $+$ $3O_2$

`=>` `m_(O_2` `=` `7,2(g)`

`n_(O_2` `=` `(7,2)/(32)=0,225(mol)`

`V_(O_2` `=` `0,225.22,4=5,04(l)`

$b)$

Theo PT: `n_(KClO_3(pư)` `=` `n_(KCl` `=` `(2)/(3)n_(O_2` `=` `0,15(mol)`

`n_(KClO_3(dư)` `=` `(19,6)/(122,5)-0,15=0,01(mol)`

`m_(KClO_3(dư)` `=` `0,01.122,5=1,225(g)`

`m_(KCl` `=` `0,15.74,5=11,175(g)`

`%m_(KClO_3(dư)` `=` `(1,225)/(11,175+1,225).100%≈9,88%`

`%m_(KCl` `=` `100-9,88=90,12%`