Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0}2A{l_2}{O_3}\\
b)\\
{n_{Al}} = \dfrac{{10,8}}{{27}} = 0,4\,mol\\
{n_{{O_2}}} = \dfrac{{0,4 \times 3}}{4} = 0,3\,mol\\
{V_{{O_2}}} = 0,3 \times 22,4 = 6,72l\\
c)\\
{n_{A{l_2}{O_3}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{m_{A{l_2}{O_3}}} = 0,2 \times 102 = 20,4g\\
2)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
b)\\
{n_{{O_2}}} = \dfrac{{3,6}}{{22,4}} = \dfrac{9}{{56}}\,mol\\
{n_{Al}} = \dfrac{{\dfrac{9}{{56}} \times 4}}{3} = \dfrac{3}{{14}}\,mol\\
{m_{Al}} = \dfrac{3}{{14}} \times 27 = 5,79(g)\\
c)\\
{n_{A{l_2}{O_3}}} = \dfrac{{\dfrac{3}{{14}}}}{2} = \dfrac{3}{{28}}\,mol\\
{m_{A{l_2}{O_3}}} = \dfrac{3}{{28}} \times 102 = 10,93g\\
3)\\
a)\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
b)\\
{n_{Mg}} = \dfrac{{9,6}}{{24}} = 0,4\,mol\\
{n_{{O_2}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{V_{{O_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{MgO}} = {n_{Mg}} = 0,4\,mol\\
{m_{MgO}} = 0,4 \times 40 = 16g\\
4)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
b)\\
{n_{Al}} = \dfrac{{21,6}}{{27}} = 0,8\,mol\\
{n_{{O_2}}} = \dfrac{{0,8 \times 3}}{4} = 0,6mol\\
{V_{{O_2}}} = 0,6 \times 22,4 = 13,44l\\
c)\\
{n_{A{l_2}{O_3}}} = \dfrac{{0,8}}{2} = 0,4\,mol\\
{m_{A{l_2}{O_3}}} = 0,4 \times 102 = 40,8g
\end{array}\)

Đáp án:

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