Đáp án:

\[m=5,67\ \rm{gam}\]

Giải thích các bước giải:

\[\underbrace {\left\{ \begin{array}{l} Al\\ A{l_2}{O_3}\\ Al{(OH)_3} \end{array} \right\}}_{m\;{\rm{gam}}}+\xrightarrow[19,6\%]{H_2SO_4}\left\langle \begin{array}{l} A{l_2}{(S{O_4})_3}\\ {H_2} \end{array} \right.\] \[\begin{array}{l} {n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15\;{\rm{mol}}\\ {n_{A{l_2}{{(S{O_4})}_3}}} = \frac{{80,37 \cdot 21,302\% }}{{342}} \approx 0,05\;{\rm{mol}} \end{array}\] \[\begin{array}{l} {\rm{BTNT}}\;S:\\ {n_{{H_2}S{O_4}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}}\\ \Rightarrow {n_{{H_2}S{O_4}}} = 3 \cdot 0,05 = 0,15\;{\rm{mol}}\\ \Rightarrow {m_{dd\;{H_2}S{O_4}}} = \frac{{0,15 \cdot 98}}{{19,6\% }} = 75\;{\rm{gam}} \end{array}\]

BTKL, ta có:

\(m+ {m_{dd\;{H_2}S{O_4}}} =m_Y+m_{H_2}\) \[ \Rightarrow m = 80,37 + 0,15 \cdot 2 - 75 = 5,67\;{\rm{gam}}\]