`n_(CuO)=20/80=0.25(mol)`

`n_(PbO)=111.5/223=0.5(mol)`

`PTHH:CuO+H_2->Cu+H_2O(1)`

`PbO+H_2->Pb+H_2O(2)`

`Theo pt (1):n_(H_2)=n_(CuO)=0.25(mol)`

`Theo pt (2);n_(H_2)=n_(PbO)=0.5(mol)`

`V_(H_2)=22.4·(0.25+0.5)=16.8(l)`

Đáp án:

\({V_{{H_2}}} = 16,8{\text{ lít}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

\(PbO + {H_2}\xrightarrow{{{t^o}}}Pb + {H_2}O\)

Ta có:

\({n_{CuO}} = \frac{{20}}{{64 + 16}} = 0,25{\text{ mol}}\)

\({n_{PbO}} = \frac{{111,5}}{{207 + 16}} = 0,5{\text{ mol}}\)

\( \to {n_{{H_2}}} = {n_{CuO}} + {n_{PbO}} = 0,25 + 0,5 = 0,75{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,75.22,4 = 16,8{\text{ lít}}\)