38)

Phản ứng xảy ra:

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

Ta có: 

\({n_{Zn}} = \frac{{15,6}}{{65}} = 0,24{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{39,2}}{{98}} = 0,4{\text{mol > }}{{\text{n}}_{Zn}}\) nên axit dư.

\( \to {n_{{H_2}}} = {n_{Zn}} = 0,24{\text{ mol}} \to {{\text{n}}_{{H_2}{\text{thu được}}}} = 0,24.95\%  = 0,228{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,228.22,4 = 5,1072{\text{ lít}}\)

\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,4 - 0,24 = 0,16{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}{\text{ dư}}}} = 0,16.98 = 15,68{\text{ gam}}\)

39)

1)

Gọi số mol Zn và Fe trong hỗn hợp lần lượt là x, y.

\( \to 65x + 56y = 43,7{\text{ gam}}\)

Cho hỗn hợp kim loại tác dụng với HCl

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

\( \to {n_{{H_2}}} = x + y = \frac{{15,68}}{{22,4}} = 0,7{\text{ mol}}\)

Giải được: x=0,5; y=0,2.

\( \to {m_{Zn}} = 0,5.65 = 32,5{\text{ gam;}}{{\text{m}}_{Fe}} = 0,2.56 = 11,2{\text{ gam}}\)

Khử oxit sắt bằng khí hidro

\(F{e_3}{O_4} + 4{H_2}\xrightarrow{{{t^o}}}3Fe + 4{H_2}O\)

\({n_{F{e_3}{O_4}}} = \frac{{46,4}}{{56.3 + 16.4}} = 0,2{\text{ mol > }}\frac{1}{4}{n_{{H_2}}} \to {n_{Fe}} = \frac{3}{4}{n_{{H_2}}} = 0,525{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,525.56 = 29,4{\text{ gam}}\)

2)

Phản ứng xảy ra:

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

\({n_{Al}} = \frac{a}{{27}} \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = \frac{a}{{18}}\)

\({n_{Zn}} = {n_{{H_2}}} = \frac{b}{{65}} = \frac{a}{{18}} \to \frac{a}{b} = \frac{{18}}{{65}}\)