17)
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{6}{{24}} = 0,25{\text{ mol = }}{{\text{n}}_{{H_2}}} \to {V_{{H_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)
Dùng lượng khí hidro này để khử oxit sắt
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
Ta có: \({n_{{H_2}}} < 3{n_{F{e_2}{O_3}}} \to Fe{\text{ dư}}\)
\( \to {n_{Fe}} = \frac{2}{3}{n_{{H_2}}} = \frac{{0,5}}{3} \to {m_{Fe}} = \frac{{0,5}}{3}.56 = 9,333{\text{ gam}}\)
18)
Ta có:
\({m_{Fe}} = 14,9.56,37\% = 8,4{\text{ gam}} \to {{\text{m}}_{Zn}} = 14,9 - 8,4 = 6,5{\text{ gam}}\)
\( \to {n_{Fe}} = \frac{{8,4}}{{56}} = 0,15{\text{ mol; }}{{\text{n}}_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol}}\)
Phản ứng xảy ra:
\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
\( \to {n_{{H_2}}} = {n_{Zn}} + {n_{Fe}} = 0,25{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,15{\text{ mol; }}{{\text{n}}_{ZnS{O_4}}} = {n_{Zn}} = 0,1{\text{ mol}} \to {{\text{m}}_{FeS{O_4}}} = 0,15.(56 + 96) = 22,8{\text{ gam; }}{{\text{m}}_{ZnS{O_4}}} = 0,1.(65 + 96) = 16,1{\text{ gam}}\)
