Em tham khảo nha :
\(\begin{array}{l}
1)\\
2Ca + {O_2} \xrightarrow{t^0} 2CaO\\
{n_{Ca}} = \dfrac{m}{M} = \frac{{16}}{{40}} = 0,4mol\\
{n_{{O_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
\dfrac{{0,4}}{2} < \dfrac{{0,5}}{1} \Rightarrow {O_2}\text{ dư}\\
{n_{{O_2}d}} = {n_{{O_2}}} - \dfrac{{{n_{Ca}}}}{2} = 0,5 - \dfrac{{0,4}}{2} = 0,3mol\\
{m_{{O_2}d}} = n \times M = 0,3 \times 32 = 9,6g\\
{n_{CaO}} = {n_{Ca}} = 0,4mol\\
{m_{CaO}} = n \times M = 0,4 \times 56 = 22,4g\\
2)\\
2KMn{O_4} \xrightarrow{t^0} Mn{O_2} + {K_2}Mn{O_4} + {O_2}\\
{n_{KMn{O_4}}} = \dfrac{m}{M} = \dfrac{{23,7}}{{158}} = 0,15mol\\
{n_{{O_2}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = 0,075mol\\
{m_{{O_2}}} = n \times M = 0,075 \times 32 = 2,4g\\
H = \dfrac{{1,92}}{{2,4}} \times 100\% = 80\% \\
3)\\
2KMn{O_4} \xrightarrow{t^0} Mn{O_2} + {K_2}Mn{O_4} + {O_2}\\
{n_{{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,5mol\\
{m_{KMn{O_4}}} = n \times M = 0,5 \times 158 = 79g\\
H = 96\% \\
\Rightarrow {m_{KMn{O_4}}}\text{ cần dùng} = \dfrac{{79 \times 100}}{{96}} = 82,3g
\end{array}\)
