Đáp án:
\(\begin{array}{l}
{m_{MgO}} = 4g\\
{m_{{O_2}}} = 1,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = \dfrac{m}{M} = \dfrac{{31,6}}{{158}} = 0,2\,mol\\
{n_{{O_2}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{Mg}} = \dfrac{m}{M} = \frac{{2,4}}{{24}} = 0,1\,mol\\
2Mg + {O_2} \to 2MgO\\
\dfrac{{0,1}}{2} < \dfrac{{0,1}}{1} \Rightarrow \text{ $O_2$ dư}\\
{n_{MgO}} = {n_{Mg}} = 0,1\,mol\\
{n_{{O_2}}} = 0,1 - 0,05 = 0,05\,mol\\
{m_{MgO}} = n \times M = 0,1 \times 40 = 4g\\
{m_{{O_2}}} = n \times M = 0,05 \times 32 = 1,6g
\end{array}\)