Em tham khảo nha :

\(\begin{array}{l}
4)\\
a)\\
2Mg + {O_2} \to 2MgO\\
{n_{Mg}} = \dfrac{6}{{24}} = 0,25mol\\
{n_{MgO}} = {n_{Mg}} = 0,25mol\\
{m_{MgO}} = 0,25 \times 40 = 10g\\
b)\\
{n_{{O_2}}} = \dfrac{{{n_{Mg}}}}{2} = 0,125mol\\
{V_{{O_2}}} = 0,125 \times 22,4 = 2,8l\\
5)\\
a)\\
4Na + {O_2} \to 2N{a_2}O\\
{n_{{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Na}} = 4{n_{{O_2}}} = 1,2mol\\
{m_{Na}} = 1,2 \times 23 = 27,6g\\
b)\\
{n_{N{a_2}O}} = 2{n_{{O_2}}} = 0,6mol\\
{m_{N{a_2}O}} = 0,6 \times 62 = 37,2g\\
6)\\
a)\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{F{e_3}{O_4}}} = \dfrac{{6,96}}{{232}} = 0,03mol\\
{n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,09mol\\
{m_{Fe}} = 0,09 \times 56 = 5,04g\\
{n_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,06mol\\
{V_{{O_2}}} = 0,06 \times 22,4 = 1,344l\\
7)\\
a)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{{14,7}}{{122,5}} = 0,12mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,18mol\\
{V_{{O_2}}} = 0,18 \times 22,4 = 4,032l\\
b)\\
{n_{KCl}} = {n_{KCl{O_3}}} = 0,12mol\\
{m_{KCl}} = 0,12 \times 74,5 = 8,94g
\end{array}\)

Câu 5