Đáp án:

 \(V=6,72 lít\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

Ta có: 

\({n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol}}\)

\( \to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,3{\text{ mol}}\)

\( \to V = {V_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)

\({m_{{H_2}S{O_4}}} = 0,3.98 = 29,4{\text{ gam}}\)

\( \to {m_{dd{{\text{H}}_2}S{O_4}}} = \frac{{29,4}}{{19,6\% }} = 150{\text{ gam}}\)

BTKL:

\({m_{Al}} + {m_{dd{{\text{H}}_2}S{O_4}}} = {m_{dd\;{\text{X}}}} + {m_{{H_2}}}\)

\( \to {m_{dd{\text{ X}}}} = 5,4 + 150 - 0,3.2 = 154,6{\text{gam}}\)

\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,1{\text{ mol}}\)

\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,1.(27.2 + 96.3) = 34,2{\text{ gam}}\)

\( \to \% {m_{A{l_2}{{(S{O_4})}_3}}} = \frac{{34,2}}{{154,6}} = 22,12\% \)