Em tham khảo nha :
\(\begin{array}{l}
5)\\
2Mg + {O_2} \to 2MgO\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
hh:Mg(a\,mol),Al(b\,mol)\\
a - 2b = 0(1)\\
24a + 27b = 15(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,4mol;b = 0,2mol\\
{n_{{O_2}}} = \frac{{{n_{Mg}}}}{2} + \frac{{3{n_{Al}}}}{4} = 0,35mol\\
{V_{{O_2}}} = 0,35 \times 22,4 = 7,84l\\
{n_{MgO}} = {n_{Mg}} = 0,4mol\\
{m_{MgO}} = 0,4 \times 40 = 16g\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{m_{A{l_2}{O_3}}} = 0,1 \times 102 = 10,2g\\
m = {m_{MgO}} + {m_{A{l_2}{O_3}}} = 16 + 10,2 = 26,2g\\
\% MgO = \dfrac{{16}}{{26,2}} \times 100\% = 61,1\% \\
\% A{l_2}{O_3} = 100 - 61,1 = 38,9\% \\
6)\\
4Na + {O_2} \to 2N{a_2}O\\
2Mg + {O_2} \to 2MgO\\
{n_{{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Na(a\,mol),Mg(b\,mol)\\
a - b = 0(1)\\
\dfrac{1}{4}a + \dfrac{1}{2}b = 0,3(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,4;b = 0,4\\
{m_{Na}} = 0,4 \times 23 = 9,2g\\
{m_{Mg}} = 0,4 \times 24 = 9,6g\\
m = {m_{Na}} + {m_{Mg}} = 9,2 + 9,6 = 18,8g\\
{n_{N{a_2}O}} = \dfrac{{{n_{Na}}}}{2} = 0,2mol\\
{m_{N{a_2}O}} = 0,2 \times 62 = 12,4g\\
{n_{MgO}} = {n_{Mg}} = 0,4mol\\
{m_{MgO}} = 0,4 \times 40 = 16g\\
{m_1} = {m_{N{a_2}O}} + {m_{MgO}} = 12,4 + 16 = 28,4g
\end{array}\)
