Đáp án:
a) \(\% {m_{A{l_2}{S_3}}} = 80,65\% ;\% {m_{Al{\text{ dư}}}} = 19,35\% \)
b) \({d_{B/{H_2}}} = 10,6\)
c) m=30,72 gam
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 3S\xrightarrow{{{t^o}}}A{l_2}{S_3}\)
Ta có:
\({n_{Al}} = \frac{{10,8}}{{27}} = 0,4{\text{ mol;}}{{\text{n}}_S} = \frac{{11,52}}{{32}} = 0,36{\text{ mol}} \to {{\text{n}}_{Al}} > \frac{2}{3}{n_S}\) nên Al dư.
Ta có:
\({n_{A{l_2}{S_3}}} = \frac{1}{3}{n_S} = 0,12{\text{ mol;}}{{\text{n}}_{Al{\text{ du}}}} = 0,4 - 0,12.2 = 0,16{\text{ mol}}\)
\( \to {m_{A{l_2}{S_3}}} = 0,12.(27.2 + 32.3) = 18{\text{ gam;}}{{\text{m}}_{Al}} = 0,16.27 = 4,32{\text{ gam}}\)
\( \to \% {m_{A{l_2}{S_3}}} = \frac{{18}}{{18 + 4,32}} = 80,65\% \to \% {m_{Al{\text{ dư}}}} = 100\% - 80,65\% = 19,35\% \)
\({n_{{H_2}S}} = 3{n_{A{l_2}{S_3}}} = 0,36{\text{ mol;}}{{\text{n}}_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,24{\text{ mol}} \to {{\text{n}}_B} = 0,36 + 0,24 = 0,6{\text{ mol;}}{{\text{m}}_B} = 0,36.34 + 0,24.2 = 12,72 \to {M_B} = \frac{{12,72}}{{0,6}} = 21,2 \to {d_{B/{H_2}}} = \frac{{21,2}}{2} = 10,6\)
Dẫn hỗn hợp khí qua đồng nitrat
\(Cu{(N{O_3})_2} + {H_2}S\xrightarrow{{}}CuS + 2HN{O_3}\)
\({n_{CuS}} = {n_{{H_2}S}} = 0,32{\text{ mol}} \to {\text{m = 0}}{\text{,32}}{\text{.(64 + 32) = 30}}{\text{,72 gam}}\)
