Đáp án:
\(\begin{array}{l}
b)\\
{m_{{P_2}{O_5}}} = 2,84g\\
c)\\
{m_{KCl{O_3}}} = 4,083g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
b)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{{P_2}{O_5}}} = \dfrac{2}{5}{n_{{O_2}}} = 0,02mol\\
{m_{{P_2}{O_5}}} = n \times M = 0,02 \times 142 = 2,84g\\
c)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = \dfrac{1}{{30}}mol\\
{m_{KCl{O_3}}} = n \times M = \frac{1}{{30}} \times 122,5 = 4,083g
\end{array}\)