Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{{H_2}}} = 0,2 \times \dfrac{3}{2} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,2\,mol\\
{m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{C_M}HCl = \dfrac{{0,6}}{{0,15}} = 4M\\
d)\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{CuO}} = \dfrac{{25}}{{80}} = 0,3125mol\\
{n_{CuO}} > {n_{{H_2}}} \Rightarrow CuO \text{ dư } \\
{n_{CuO}}  \text{ dư }= 0,3125 - 0,3 = 0,0125\,mol\\
{n_{Cu}} = {n_{{H_2}}} = 0,3\,mol\\
x = 0,0125 \times 80 + 0,3 \times 64 = 20,2g
\end{array}\)