Em tham khảo nha:

\(\begin{array}{l}
1)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = 0,5 \times 1 = 0,5\,mol\\
\dfrac{{{n_{Al}}}}{2} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow \text{ $H_2SO_4$ dư } \\
{n_{{H_2}}} = 0,2 \times \dfrac{3}{2} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
m = 5,4 - 0,3 \times 2 = 4,8g\\
2)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
\text{ Gọi a là số mol Zn }\\
{n_{{H_2}}} = {n_{Zn}} = a\,mol\\
 \Rightarrow 65a - 2a = 6,3 \Rightarrow a = 0,1\,mol\\
{m_{Zn}} = 0,1 \times 65 = 6,5g \Rightarrow m = 6,5\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l \Rightarrow V = 4,48\\
{n_{HCl}} = 0,2\, \times 130\%  = 0,26mol\\
{C_M}HCl = \dfrac{{0,26}}{{0,3}} = 0,87M \Rightarrow x = 0,87
\end{array}\)