Em tham khảo nha :

\(\begin{array}{l}
24)\\
a)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
{n_{{O_2}}} = \dfrac{2}{3}{n_{Fe}} = \frac{1}{3}mol\\
{m_{{O_2}}} = n \times M = \frac{1}{3} \times 32 = 10,7g\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{{O_2}}} = \dfrac{3}{4}{n_{Al}} = 0,9375mol\\
{m_{{O_2}}} = n \times M = 0,9375 \times 32 = 30g\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
{n_{{O_2}}} = \dfrac{{{n_{Zn}}}}{2} = 0,75mol\\
{m_{{O_2}}} = n \times M = 0,75 \times 32 = 24g\\
b)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_P} = \dfrac{{3,1}}{{31}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{5}{4}{n_P} = 0,125mol\\
{m_{{O_2}}} = n \times M = 0,125 \times 32 = 4g\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
{n_S} = \dfrac{{6,4}}{{32}} = 0,2mol\\
{n_{{O_2}}} = {n_S} = 0,2mol\\
{m_{{O_2}}} = n \times M = 0,2 \times 32 = 6,4g\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
{n_C} = \dfrac{{3,6}}{{12}} = 0,3mol\\
{n_{{O_2}}} = {n_C} = 0,3mol\\
{m_{{O_2}}} = n \times M = 0,3 \times 32 = 9,6g\\
c)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{n_{C{H_4}}} = \dfrac{{1,6}}{{16}} = 0,1mol\\
{n_{{O_2}}} = 2{n_{C{H_4}}} = 0,2mol\\
{m_{{O_2}}} = n \times M = 0,2 \times 32 = 6,4g\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}\\
{n_{CO}} = \dfrac{{2,8}}{{28}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{{{n_{CO}}}}{2} = 0,05mol\\
{m_{{O_2}}} = n \times M = 0,05 \times 32 = 1,6g\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
{n_{{C_4}{H_{10}}}} = \dfrac{{0,58}}{{58}} = 0,01mol\\
{n_{{O_2}}} = \dfrac{{13}}{2}{n_{{C_4}{H_{10}}}} = 0,065mol\\
{m_{{O_2}}} = n \times M = 0,065 \times 32 = 2,08g\\
25)\\
a)\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O(1)\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}(2)\\
b)\\
{n_{C{O_2}}} = \dfrac{{8,8}}{{44}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{{9,6}}{{32}} = 0,3mol\\
{n_{{O_2}(2)}} = \dfrac{{{n_{C{O_2}}}}}{2} = 0,1mol\\
{n_{{O_2}(1)}} = 0,3 - 0,1 = 0,2mol\\
{n_{{H_2}}} = 2{n_{{O_2}}} = 0,4mol\\
\% {n_{{H_2}}} = \dfrac{{0,4}}{{0,4 + 0,2}} \times 100\%  = 66,7\% \\
\% {n_{CO}} = 100 - 66,7 = 33,3\% \\
{m_{CO}} = 0,1 \times 28 = 2,8g\\
{m_{{H_2}}} = 0,4 \times 2 = 0,8g\\
\% {m_{{H_2}}} = \dfrac{{0,8}}{{0,8 + 2,8}} \times 100\%  = 22,2\% \\
\% {m_{CO}} = 100 - 22,2 = 77,8\% 
\end{array}\)

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