Em tham khảo nha :

\(\begin{array}{l}
a)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
{m_{KCl{O_3}}} = \dfrac{{130 \times 91}}{{100}} = 118,3g\\
{n_{KCl{O_3}}} = \dfrac{{118,3}}{{122,5}} = 0,966mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 1,449mol\\
{V_{{O_2}}} = 1,449 \times 22,4 = 32,4576l\\
b)\\
2Mg + {O_2} \to 2MgO\\
{n_{{O_2}}} = \dfrac{{1,449}}{2} = 0,7245mol\\
{n_{Mg}} = 2{n_{{O_2}}} = 1,449mol\\
{m_{Mg}} = 1,449 \times 24 = 34,776g\\
{n_{MgO}} = {n_{Mg}} = 1,449mol\\
{m_{MgO}} = 1,449 \times 40 = 57,96g
\end{array}\)

Đáp án:

$\downarrow$

Giải thích các bước giải:

a, `-` PTHH: `2KClO_3->2KCl+3O_2`

* `m_{KClO_3}=130xx91%=118,3(gam)`

`->n_{KClO_3}=(118,3)/(122,5)=0,966(mol)`

Theo PTHH `->n_{O_2}=3/2 n_{KClO_3}=1,449(mol)`

`->V_{O_2(đktc)}=1,449xx22,4=32,4576(lít)`

b, * `n_{O_2}=(1,449)/2=0,7245(mol)`

`-` PTHH: `2Mg+O_2->2MgO`

Theo PTHH `->n_{Mg}=2n_{O_2}=1,449(mol)`

`->m_{Mg}=1,449xx24=34,776(gam)`

Ta thấy: `n_{Mg}=n_{MgO}=1,449(mol)`

`->m_{MgO}=1,449xx40=57,96(gam)`

$\text{#Kudo}$