Đáp án:
\({m_{A{l_2}{{(S{O_4})}_3}}} = 68,4{\text{ gam}}\)
\({m_{{H_2}S{O_4}}}= 39,2{\text{ gam}}\)
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:
\({n_{Al}} = \frac{{10,8}}{{27}} = 0,4{\text{ }}mol\)
\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = \frac{1}{2}.0,4 = 0,2{\text{ mol}}\)
\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,2.(27.2 + 96.3) = 68,4{\text{ gam}}\)
\({n_{{H_2}}} = \frac{{0,8}}{2} = 0,4{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,4.98 = 39,2{\text{ gam}}\)