Đáp án:
\( a = 2,7{\text{ gam}}\)
\(b = 17,1{\text{ gam}}\)
\({m_{{H_2}S{O_4}}} = 14,7{\text{ gam}}\)
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:
\({n_{{H_2}}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol}}\)
Theo phản ứng:
\({n_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,1{\text{ mol;}}{{\text{n}}_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{3}{n_{{H_2}}} = 0,0{\text{ mol}}\)
\( \to a = {m_{Al}} = 0,1.27 = 2,7{\text{ gam}}\)
\(b = {m_{A{l_2}{{(S{O_4})}_3}}} = 0,05.(27.2 + 96.3) = 17,1{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,15{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,15.98 = 14,7{\text{ gam}}\)
