1)
Phản ứng xảy ra:
\(KCl{O_3} + 6HCl\xrightarrow{{}}KCl + 3C{l_2} + 3{H_2}O\)
Ta có:
\({n_{KCl{O_3}}} = \frac{{14,7}}{{39 + 35,5 + 16.3}} = 0,12{\text{ mol}}\)
\( \to {n_{C{l_2}}} = 3{n_{KCl{O_3}}} = 0,36{\text{ mol}}\)
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
\( \to {n_{{O_2}{\text{ lt}}}} = \frac{3}{2}{n_{KCl{O_3}}} = 0,18{\text{ mol}}\)
\( \to {n_{{O_2}}} = 0,18.80\% = 0,144{\text{ mol}}\)
\( \to {V_1}:{V_2} = 0,36:0,144 = 5:2\)
2)
Phản ứng xảy ra:
\(FeS + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}S\)
\({H_2}S + Pb{(N{O_3})_2}\xrightarrow{{}}PbS + 2HN{O_3}\)
Ta có:
\({m_{FeS}} = 12,8.\frac{{11}}{{11 + 9}} = 7,04{\text{ gam}}\)
\( \to {n_{FeS}} = \frac{{7,04}}{{56 + 32}} = 0,08{\text{ mol = }}{{\text{n}}_{{H_2}S}}\)
\( \to {n_{PbS}} = {n_{{H_2}S}} = 0,08{\text{ mol}}\)
\( \to m= {m_{PbS}} = 0,08.(207 + 32) = 19,12{\text{ gam}}\)