Đáp án:
\( {m_{Al}} = 5,4{\text{ gam}}; {{\text{m}}_{A{l_2}{O_3}}} = 10,2{\text{ gam}}\)
\( {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = 600{\text{ gam}}\)
Giải thích các bước giải:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
\(A{l_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}O\)
\( \to {n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}} \to {{\text{n}}_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,2{\text{ mol}}\)
\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}} \to {{\text{m}}_{A{l_2}{O_3}}} = 15,6 - 5,4 = 10,2{\text{ gam}}\)
\( \to {n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol;}}{{\text{n}}_{A{l_2}{O_3}}} = \frac{{10,2}}{{27.2 + 16.3}} = 0,1{\text{ mol}}\)
\({n_{{H_2}S{O_4}}} = \frac{3}{2}{n_{Al}} + 3n_{Al_2O_3} = \frac{3}{2}.0,2 + 0,1.3 = 0,6{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,6.98 = 58,8{\text{ gam}}\)
\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{58,8}}{{9,8\% }} = 600{\text{ gam}}\)
