Đáp án:

 38 bình \(H_2\)

\(Al\) (nhôm)

\({m_{A{l_2}{{(S{O_4})}_3}}} = 13,68{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2{\text{A}} + 3{H_2}S{O_4}\xrightarrow{{}}{A_2}{(S{O_4})_3} + 3{H_2}\)

Ta có:

\({n_{{H_2}S{O_4}}} = \frac{{11,76}}{{98}} = 0,12{\text{ mol = }}{{\text{n}}_{{H_2}}}\)

\( \to {V_{{H_2}{\text{ lt}}}} = 0,12.22,4 = 2,688{\text{ lit}} \to {{\text{V}}_{{H_2}}} = 2,688.70\%  = 1,8816{\text{ lít = 1881,6 ml}}\)

Số bình \(H_2\)\( = \frac{{1881,6}}{{50}} = 37,632 \approx 38{\text{ }}\)

\( \to {n_A} = \frac{2}{3}{n_{{H_2}S{O_4}}} = 0,08{\text{ mol}}\)

\( \to {M_A} = \frac{{2,16}}{{0,08}} = 27 \to A:Al\)

\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,04{\text{ mol}}\)

\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,04.(27.2 + 96.3 = 13,68{\text{ gam}}\)