Đáp án + Giải thích các bước giải:

$a\big)$

$n_{H_2}=\frac{1,12}{22,4}=0,05(mol)$

$CuO+CO\xrightarrow{t^o}Cu+CO_2(1)$

$Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2(2)$

$Fe+2HCl\to FeCl_2+H_2(3)$

Theo PT(3): $n_{Fe}=n_{H_2}=0,05(mol)$

Theo PT(2): $n_{Fe_2O_3}=\frac{1}{2}n_{Fe}=0,025(mol)$

$\to m_{Fe_2O_3}=0,025.160=4(g)$

$\to m_{CuO}=5,2-4=1,2(g)$

$b\big)$

$n_{CuO}=\frac{1,2}{80}=0,015(mol)$

Theo PT: `\sum n_{CO}=n_{CuO}+3n_{Fe_2O_3}=0,09(mol)`

$\to V_{CO(25^oC,1\,bar)}=0,09.24,79=2,2311(l)$

Đáp án:

\({m_{F{{\text{e}}_2}{O_3}}} = 4{\text{ gam}}; {{\text{m}}_{CuO}} = 1,2{\text{ gam}}\)

\({V_{CO}} = 2,2311{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(CuO + CO\xrightarrow{{{t^o}}}Cu + C{O_2}\)

\(F{{\text{e}}_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2F{\text{e}} + 3{H_2}O\)

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\({n_{{H_2}}} = {n_{F{\text{e}}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol}} \to {{\text{n}}_{F{{\text{e}}_2}{O_3}}} = \frac{1}{2}{n_{F{\text{e}}}} = 0,025{\text{ mol}}\)

\( \to {m_{F{{\text{e}}_2}{O_3}}} = 0,025.160 = 4{\text{ gam}} \to {{\text{m}}_{CuO}} = 1,2{\text{ gam}}\)

\( \to {n_{CuO}} = \frac{{1,2}}{{64 + 16}} = 0,015{\text{ mol}} \to {{\text{n}}_{CO}} = 3{n_{F{{\text{e}}_2}{O_3}}} + {n_{CuO}} = 0,025.3 + 0,015 = 0,09{\text{ mol}}\)

\( \to {V_{CO}} = 0,09.24,79 = 2,2311{\text{ lít}}\)