Đáp án:
\( a = 18{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
\(A{l_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}\frac{3}{2}{n_{Al}}\)
\( \to {n_{Al}} = 0,1{\text{ mol}}\)
\({n_{{H_2}S{O_4}}} = 0,3.2 = 0,6{\text{ mol = }}\frac{3}{2}{n_{Al}} + 3{n_{A{l_2}{O_3}}}\)
\( \to {n_{A{l_2}{O_3}}} = 0,15{\text{ mol}}\)
\( \to a = {m_{Al}} + {m_{A{l_2}{O_3}}} = 0,1.27 + 0,15.(27.2 + 16.3) = 18{\text{ gam}}\)
