Đáp án:
\({V_{{H_2}}} = 11,2{\text{ lít}}\)
\(\% {m_{Zn}} = 95\%; \% {m_{Al}} = 5\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
Ta có:
\({n_{HCl}} = \frac{{16,425}}{{36,5}} = 0,45{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{26,95}}{{98}} = 0,275{\text{ mol}}\)
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} + {n_{{H_2}S{O_4}}} = \frac{1}{2}.0,45 + 0,275 = 0,5{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,5.22,4 = 11,2{\text{ lít}}\)
Gọi số mol \(Zn;Al\) lần lượt là \(x;y\)
\( \to 65x + 27y = 28,7{\text{ gam}}\)
\({n_{{H_2}}} = {n_{Zn}} + \frac{3}{2}{n_{Al}} = x + 1,5y = 0,5\)
Giải được:
\(x = \frac{{197}}{{470}};y = \frac{{38}}{{705}}\)
\( \to {m_{Zn}} = 65x = 27,25{\text{ gam}}\)
\( \to \% {m_{Zn}} = \frac{{27,25}}{{28,7}} = 95\% \to \% {m_{Al}} = 5\% \)
