Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
2X + 2nHCl \to 2XC{l_n} + n{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_X} = \dfrac{{0,3 \times 2}}{n} = \dfrac{{0,6}}{n}\,mol\\
{M_X} = \dfrac{{5,4}}{{\dfrac{{0,6}}{n}}} = 9n\,g/mol\\
n = 3 \Rightarrow {M_X} = 27g/mol \Rightarrow X:Al\\
b)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{V_{{\rm{ddHCl}}}} = \dfrac{{0,6}}{3} = 0,2l\\
{n_{AlC{l_3}}} = 0,3 \times \dfrac{2}{3} = 0,2\,mol\\
{C_M}AlC{l_3} = \dfrac{{0,2}}{{0,2}} = 1M\\
5)\\
a)\\
{X_2}{O_3} + 3{H_2} \to 2X + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
{n_{{X_2}{O_3}}} = \dfrac{{0,6}}{3} = 0,2\,mol\\
{M_{{X_2}{O_3}}} = \dfrac{{32}}{{0,2}} = 160g/mol \Rightarrow 2{M_X} + 16 \times 3 = 160\\
 \Rightarrow {M_X} = 56g/mol \Rightarrow X:Fe\\
b)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{HCl}} = 6{n_{F{e_2}{O_3}}} = 1,2\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{1,2}}{{2,5}} = 0,48l\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,4\,mol\\
{m_{FeC{l_3}}} = 0,4 \times 162,5 = 65g\\
9)\\
a)\\
{A_2}{(S{O_4})_n} + nBaC{l_2} \to nBaS{O_4} + 2AC{l_n}\\
{n_{BaS{O_4}}} = \dfrac{{46,6}}{{233}} = 0,2\,mol\\
{n_{{A_2}{{(S{O_4})}_n}}} = 0,2 \times \dfrac{1}{n} = \dfrac{{0,2}}{n}\,mol\\
{M_{{A_2}{{(S{O_4})}_n}}} = \dfrac{{24}}{{\dfrac{{0,2}}{n}}} = 120n\,g/mol\\
 \Rightarrow 2{M_A} + 96n = 120n \Rightarrow {M_A} = 12n\,g/mol\\
n = 2 \Rightarrow {M_A} = 24g/mol \Rightarrow A:Mg\\
CTHH:MgS{O_4}\\
b)\\
{n_{BaC{l_2}}} = {n_{BaS{O_4}}} = 0,2\,mol\\
{V_{{\rm{dd}}BaC{l_2}}} = \dfrac{{0,2}}{2} = 0,1l
\end{array}\)