Em tham khảo nha:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
{n_{Zn}} = \dfrac{{2,6}}{{65}} = 0,04\,mol\\
{n_{Al}} = \dfrac{{0,81}}{{27}} = 0,03mol\\
{n_{{H_2}}} = 0,04 + 0,03 \times \dfrac{3}{2} = 0,085\,mol\\
{V_{{H_2}}} = 0,085 \times 22,4 = 1,904l\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,04\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,03\,mol\\
{C_M}ZnC{l_2} = \dfrac{{0,04}}{{0,04}} = 1M\\
{C_M}AlC{l_3} = \dfrac{{0,03}}{{0,04}} = 0,75M\\
{m_{{\rm{dd}}HCl}} = 40 \times 1,072 = 42,88g\\
{m_{{\rm{dd}}spu}} = 2,6 + 0,81 + 42,88 - 0,085 \times 2 = 46,12g\\
{C_\% }ZnC{l_2} = \dfrac{{0,04 \times 136}}{{42,61}} \times 100\% = 12,77\% \\
{C_\% }AlC{l_3} = \dfrac{{0,03 \times 133,5}}{{42,61}} \times 100\% = 9,4\%
\end{array}\)