Em tham khảo nha :
\(\begin{array}{l}
4)\\
{A_2}{O_3} + 3{H_2} \to 2A + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,68}}{{22,4}} = 0,075mol\\
{n_{{A_2}{O_3}}} = \dfrac{{{n_{{H_2}}}}}{3} = 0,025mol\\
{M_{{A_2}{O_3}}} = \dfrac{m}{n} = \dfrac{4}{{0,025}} = 160dvC\\
\Rightarrow 2{M_A} + 3{M_O} = 160\\
\Rightarrow {M_A} = \dfrac{{160 - 3 \times 16}}{2} = 56dvC\\
\Rightarrow A:\text{Sắt}(Fe)\\
CTHH:F{e_2}{O_3}\\
5)\\
{A_2}O + 2HCl \to 2NaCl + {H_2}O\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{10,95}}{{36,5}} = 0,3mol\\
{n_{{A_2}O}} = \dfrac{{{n_{HCl}}}}{2} = 0,15mol\\
{M_{{A_2}O}} = \dfrac{m}{n} = \dfrac{{9,3}}{{0,15}} = 62dvC\\
2{M_A} + {M_O} = 62\\
\Rightarrow {M_A} = \frac{{62 - 16}}{2} = 23dvC\\
\Rightarrow A:Natri(Na)\\
CTHH:N{a_2}O\\
{n_{NaCl}} = {n_{HCl}} = 0,3mol\\
{m_{NaCl}} = n \times M = 0,3 \times 58,5 = 17,55g
\end{array}\)
