1)
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol < }}\frac{1}{2}{n_{HCl}}\)
Do vậy \(HCl\) dư.
\( \to {n_{HCl{\text{ dư}}}} = {n_{HCl}} - 2{n_{Zn}} = 0,5 - 0,2.2 = 0,1{\text{ mol}}\)
\( \to {m_{HCl{\text{ dư}}}} = 0,1.36,5 = 3,65{\text{ gam}}\)
\({n_{{H_2}}} = {n_{Zn}} = 0,2{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
2)
Phản ứng xảy ra:
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)
Ta có:
\({m_{CuO}} = 16.25\% = 4{\text{ gam}} \to {m_{F{e_2}{O_3}}} = 12{\text{ gam}}\)
\( \to {n_{CuO}} = \frac{4}{{64 + 16}} = 0,05{\text{ mol;}}{{\text{n}}_{F{e_2}{O_3}}} = \frac{{12}}{{56.2 + 16.3}} = 0,075{\text{ mol}}\)
\( \to {n_{CuO}} = {n_{Cu}} = 0,05{\text{ mol;}}{{\text{n}}_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,075.2 = 0,15{\text{ mol}}\)
\( \to {m_{Cu}} = 0,05.64 = 3,2{\text{ gam;}}{{\text{m}}_{Fe}} = 0,15.56 = 8,4{\text{ gam}}\)
\({n_{{H_2}}} = {n_{CuO}} + 3{n_{F{e_2}{O_3}}} = 0,05 + 0,075.3 = 0,275{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,275.22,4 = 6,16{\text{ lít}}\)
