Đáp án:
\(\begin{array}{l}
a.{V_{S{O_2}}} = 8,96l\\
b.C{M_{{H_2}S{O_4}}} = 3M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}
\end{array}\)
\(\begin{array}{l}
{n_{{H_2}}} = 0,3mol\\
\to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
\to {m_{Al}} = 5,4g\\
\to {m_{Cu}} = 11,8 - 5,4 = 6,4g\\
\to {n_{Cu}} = 0,1mol
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Al}} + {n_{Cu}} = 0,4mol\\
\to {V_{S{O_2}}} = 8,96l\\
b.\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,3}}{{0,1}} = 3M
\end{array}\)
