Đáp án:

\({m_{FeO}} = 7,2{\text{gam;}}{{\text{m}}_{CuO}} = 16{\text{ gam}}\)

\(\% {m_{FeO}} = 31,03\% ;\% {m_{CuO}} = 68,97\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(FeO + {H_2}\xrightarrow{{}}Fe + {H_2}O\)

\(CuO + {H_2}\xrightarrow{{}}Cu + {H_2}O\)

Ta có: 

\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol;}}{{\text{n}}_{Fe}} = {n_{FeO}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol}} \to {{\text{n}}_{CuO}} = {n_{{H_2}}} - {n_{FeO}} = 0,2{\text{ mol}}\)

\( \to {m_{FeO}} = 0,1.(56 + 16) = 7,2{\text{gam;}}{{\text{m}}_{CuO}} = 0,2.(64 + 16) = 16{\text{ gam}}\)

\( \to \% {m_{FeO}} = \frac{{7,2}}{{7,2 + 16}} = 31,03\%  \to \% {m_{CuO}} = 68,97\% \)