Đáp án:

Giải thích các bước giải:

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

Theo PT: `nKClO_3 = nO_2 . 2/3 (mol)`

`a.`

`nO_2 = 48/32 = 1,5 (mol)`

`-> mKClO_3 = 1 . 122,5 = 122,5 (g)`

`b.`

`nO_2 = (44,8)/(22,4) = 2 (mol)`

`-> mKClO_3 = 2 . 122,5 = 245 (g)`

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{n_{{O_2}}} = \dfrac{m}{M} = \dfrac{{48}}{{32}} = 1,5\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 1,5 \times \dfrac{2}{3} = 1\,mol\\
{m_{KCl{O_3}}} = n \times M = 1 \times 122,5 = 122,5g\\
b)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{44,8}}{{22,4}} = 2\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 2 \times \dfrac{2}{3} = \dfrac{4}{3}\,mol\\
{m_{KCl{O_3}}} = n \times M = \dfrac{4}{3} \times 122,5 = 163,33g
\end{array}\)