Đáp án:
\(\begin{array}{l}
a)\\
{m_{Al}} = 5,4g\\
{m_{Mg}} = 2,4g\\
b)\\
{m_{{\rm{dd}}HCl}} = 800g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
27a + 24b = 7,8\\
1,5a + b = 0,4
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,8\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,8 \times 36,5}}{{3,65\% }} = 800g
\end{array}\)
