Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
 \to {m_{Al}} = 5,4g\\
 \to {m_{A{l_2}{O_3}}} = 15,6 - 5,4 = 10,2g\\
 \to {n_{A{l_2}{O_3}}} = 0,1mol\\
 \to {n_{HCl}} = 3{n_{Al}} + 6{n_{A{l_2}{O_3}}} = 1,2mol\\
 \to {m_{HCl}} = 43,8g\\
 \to {m_{HCl}}{\rm{dd}} = \dfrac{{43,8 \times 100}}{{14,6}} = 300g\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
 \to {n_{S{O_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
 \to {V_{S{O_2}}} = 6,72l\\
{n_{{H_2}S{O_4}}} = 3({n_{Al}} + {n_{A{l_2}{O_3}}}) = 0,9mol\\
 \to {m_{{H_2}S{O_4}}} = 88,2g\\
 \to {m_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{88,2 \times 100}}{{95}} = 92,84g\\
 \to {V_{{H_2}S{O_4}}} = \dfrac{{92,84}}{{1,75}} = 53,05ml
\end{array}\)