Đáp án:

\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 2,24l\\
b)\\
{m_{HC{l_d}}} = 1,825g\\
c)\\
{m_{Al}} = 1,8g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{6,5}}{{65}} = 0,1mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,25}}{2} \Rightarrow HCl\text{ dư}\\
{n_{{H_2}}} = {n_{Zn}} = 0,1mol\\
{V_{{H_2}}} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
b)\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{Zn}} = 0,25 - 0,1 \times 2 = 0,05mol\\
{m_{HC{l_d}}} = n \times M = 0,05 \times 36,5 = 1,825g\\
c)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = \dfrac{1}{{15}}mol\\
{m_{Al}} = n \times M = \dfrac{1}{{15}} \times 27 = 1,8g
\end{array}\)