Đáp án:

$\downarrow$

Giải thích các bước giải:

 Đổi `250ml=0,25l`

`->V_{O_2}=4xx3xx0,25=3(lít)`

`->n_{O_2}=5/(22,4) =15/122 (mol)`

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

Theo PTHH `-> n_{KClO_3}=2/3 n_{O_2}=5/36 (mol)`

`->m_{KClO_2}=5/36 xx122,5=10,9375(gam)`

* `2Cu+O_2` $\xrightarrow{t^o}$ `2CuO`

- `n_{Cu}=(5,6)/64=0,0875(mol)`

Ta thấy: `n_{O_2}=1/2 n_{Cu}=0,04375(mol)`

Do `V_{kk}=5V_{O_2}`

`->V_{kk}=5xx0,04375xx22,4=4,9(lít)`

$\text{#Kudo}$

Em tham khảo nha:

\(\begin{array}{l}
{V_{{O_2}}} = 4 \times 250 \times 3 = 3000ml = 3l\\
{n_{{O_2}}} = \dfrac{3}{{22,4}} = \dfrac{{15}}{{112}}\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{{15}}{{112}} \times \dfrac{2}{3} = \dfrac{5}{{56}}\,mol\\
{m_{KCl{O_3}}} = \dfrac{5}{{56}} \times 122,5 = 10,9375g\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
{n_{Cu}} = \dfrac{{5,6}}{{64}} = 0,0875\,mol\\
{n_{{O_2}}} = \dfrac{{0,0875}}{2} = 0,04375\,mol\\
{V_{kk}} = 0,04375 \times 5 \times 22,4 = 4,9l
\end{array}\)