Em tham khảo nha:

\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{9,75}}{{65}} = 0,15\,mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} + \dfrac{{{n_{Zn}}}}{2} = 2 \times 0,5 + \dfrac{{0,15}}{2} = 1,075\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{{1,075 \times 2}}{3} = \dfrac{{43}}{{60}}mol\\
{m_{KCl{O_3}}} = \dfrac{{43}}{{60}} \times 122,5 = 87,792g\\
b)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{9,75}}{{65}} = 0,15\,mol\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{6,75}}{{27}} = 0,25\,mol\\
{n_{{O_2}}} = \dfrac{{3{n_{Al}}}}{4} + \dfrac{{{n_{Zn}}}}{2} = \dfrac{{3 \times 0,25}}{4} + \dfrac{{0,15}}{2} = 0,2625\,mol\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{{0,2625 \times 2}}{3} = 0,175mol\\
{m_{KCl{O_3}}} = 0,175 \times 122,5 = 21,4375g
\end{array}\)

Đáp án:

 Trg hình nha

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