Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
hh:Al(a\,mol);Zn(b\,mol)\\
\left\{ \begin{array}{l}
27a + 65 = 3,68\\
\dfrac{3}{2}a + b = 0,1
\end{array} \right.\\
\Rightarrow a = 0,04;b = 0,04\\
{m_{Al}} = 0,04 \times 27 = 1,08g\\
{m_{Zn}} = 3,68 - 1,08 = 2,6g\\
b)\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Zn}} = 0,2mol\\
{m_{HCl}} = 0,2 \times 36,5 = 7,3g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{7,3 \times 100}}{{10}} = 73g\\
{m_{{\rm{dd}}spu}} = 3,68 + 73 - 0,1 \times 2 = 76,48g\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,04mol\\
{m_{AlC{l_3}}} = 0,04 \times 133,5 = 5,34g\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,04mol\\
{m_{ZnC{l_2}}} = 0,04 \times 136 = 5,44g\\
C{\% _{AlC{l_3}}} = \dfrac{{5,34}}{{76,48}} \times 100\% = 6,98\% \\
C{\% _{ZnC{l_2}}} = \dfrac{{5,4}}{{76,48}} \times 100\% = 7,11\%
\end{array}\)
