Em tham khảo nha:
\(\begin{array}{l}
10)\\
a)\\
{m_{Fe}} = 32 \times 70\% = 22,4g\\
{m_{Mg}} = 32 - 22,4 = 9,6g\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{22,4}}{{56}} = 0,4\,mol\\
{n_{Mg}} = \dfrac{m}{M} = \dfrac{{9,6}}{{24}} = 0,4\,mol\\
3Fe + 4{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
{n_{{O_2}}} = 0,4 \times \dfrac{4}{3} + 0,4 \times \dfrac{1}{2} = \dfrac{{11}}{{15}}mol\\
{V_{kk}} = \dfrac{{11}}{{15}} \times 5 \times 22,4 = 82,13l\\
b)\\
{n_{F{e_3}{O_4}}} = \dfrac{{{n_{Fe}}}}{3} = \dfrac{{0,4}}{3} = \dfrac{2}{{15}}\,mol\\
{n_{MgO}} = {n_{Mg}} = 0,4\,mol\\
m = \dfrac{2}{{15}} \times 232 + 0,4 \times 40 = 46,93g\\
11)\\
a)\\
{V_{C{H_4}}} = 28 \times 20\% = 5,6l\\
{V_{{C_2}{H_2}}} = 28 - 5,6 = 22,4l\\
{n_{C{H_4}}} = \dfrac{V}{{22,4}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{n_{{C_2}{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{22,4}}{{22,4}} = 1\,mol\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 1 \times \dfrac{5}{2} + 0,25 \times 2 = 3\,mol\\
{V_{kk}} = 3 \times 5 \times 22,4 = 336l\\
b)\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_2}}} + {n_{C{H_4}}} = 2,25\,mol\\
{V_{C{O_2}}} = 2,25 \times 22,4 = 50,4l
\end{array}\)
