Đáp án:
\({{\text{V}}_{{H_2}}} = 6,72{\text{ lít}}\)
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:
\({n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}}} = 0,2.2 = 0,4{\text{ mol > }}\frac{3}{2}{n_{Al}}\) nên axit dư.
Ta có:
\({n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,3{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,1{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}}} = 0,4 - \frac{3}{2}{n_{Al}} = 0,1{\text{ mol}} \to {{\text{C}}_{M{\text{ A}}{{\text{l}}_2}{{(S{O_4})}_3}}} = {C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,1}}{{0,2}} = 0,5M\)
Ta có:
\({m_{{H_2}S{O_4}{\text{ ban đầu}}}} = 0,4.98 = 39,2{\text{ gam; }}{{\text{m}}_{dd{{\text{H}}_2}S{O_4}}} = 200.1,84 = 368{\text{ gam}}\)
\( \to C{\% _{{H_2}S{O_4}}} = \frac{{39,2}}{{368}} = 10,65\% \)
BTKL:
\({m_{dd{\text{ sau phản ứng}}}} = {m_{Al}} + {m_{{H_2}S{O_4}}} - {m_{{H_2}}} = 5,4 + 368 - 0,3.2 = 372,8{\text{ gam}}\)
\({m_{A{l_2}{{(S{O_4})}_3}}} = 34,2{\text{ gam; }}{{\text{m}}_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}} \to {\text{C}}{{\text{\% }}_{A{l_2}{{(S{O_4})}_3}}} = 9,17\% ;C{\% _{{H_2}S{O_4}}} = 2,63\% \)
