Đáp án:
\({C_{{M_{HCl}}}} = 3,5M\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{1,68}}{{22,4}} = 0,075mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,075mol\\
{m_{Cu}} = 0,075 \times 64 = 4,8g\\
{n_{{H_2}}} = \dfrac{{3,92}}{{22,4}} = 0,175mol\\
hh:Al(a\,mol);Mg(b\,mol)\\
\left\{ \begin{array}{l}
27a + 24b = 3,75\\
\dfrac{3}{2}a + b = 0,175
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,1\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Mg}} = 0,35mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,35}}{{0,1}} = 3,5M
\end{array}\)