Em tham khảo nha:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{10,8}}{{27}} = 0,4\,mol\\
{n_{{H_2}}} = \dfrac{{0,4 \times 3}}{2} = 0,6\,mol\\
{V_{{H_2}}} = n \times 22,4 = 0,6 \times 22,4 = 13,44l\\
b)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,4\,mol\\
{m_{AlC{l_3}}} = n \times M = 0,4 \times 133,5 = 53,4g\\
c)\\
{n_{HCl}} = 3{n_{Al}} = 0,4 \times 3 = 1,2\,mol\\
{C_M}HCl = \dfrac{n}{V} = \dfrac{{1,2}}{{0,6}} = 2M\\
d)\\
FeO + {H_2} \xrightarrow{t^0} Fe + {H_2}O\\
{n_{FeO}} = {n_{{H_2}}} = 0,6\,mol\\
{m_{FeO}} = n \times M = 0,6 \times 72 = 43,2g
\end{array}\)