Em tham khảo nha :
\(\begin{array}{l}
1)\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
2)\\
2Ca + {O_2} \xrightarrow{t^0} 2CaO\\
{n_{Ca}} = \dfrac{{40}}{{40}} = 1mol\\
{V_{{O_2}}} = \dfrac{{{V_{kk}}}}{5} = \dfrac{{22,4}}{5} = 4,48l\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
\dfrac{1}{2} > \dfrac{{0,2}}{1} \Rightarrow Ca\text{ dư}\\
{n_{C{a_d}}} = {n_{Ca}} - 2{n_{{O_2}}} = 1 - 0,4 = 0,6mol\\
{m_{C{a_d}}} = n \times M = 0,6 \times 40 = 24g\\
3)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{{O_2}}} = \dfrac{{33,6}}{{22,4}} = 1,5mol\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = 1mol\\
{m_{KCl{O_3}}} = n \times M = 1 \times 122,5 = 122,5g\\
\text{Khối lượng $KClO_3$ cần dùng là :}\\
{m_{KCl{O_3}}} = \dfrac{{122,5 \times 100}}{{90}} = 136,11g
\end{array}\)
