Đáp án:
\(\begin{array}{l}
{V_{{H_2}}} = 13,44l\\
C{M_{{H_2}S{O_4}}} = 3M\\
{m_{{\rm{dd}}}} = 237,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a/2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b/\\
{n_{Al}} = 0,4mol\\
\to {n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,6mol\\
\to {V_{{H_2}}} = 13,44l
\end{array}\)
\(\begin{array}{l}
c/\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 0,6mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,6}}{{0,2}} = 3M\\
d/\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 200 \times 1,14 = 228g\\
{m_{{\rm{dd}}}} = {m_{Al}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 237,6g
\end{array}\)
