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:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol;}}{{\text{n}}_{Fe}} = {n_{FeO}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol}} \to {{\text{n}}_{CuO}} = {n_{{H_2}}} - {n_{FeO}} = 0,2{\text{ mol}}\)
\( \to {m_{FeO}} = 0,1.(56 + 16) = 7,2{\text{gam;}}{{\text{m}}_{CuO}} = 0,2.(64 + 16) = 16{\text{ gam}}\)
\( \to \% {m_{FeO}} = \frac{{7,2}}{{7,2 + 16}} = 31,03\% \to \% {m_{CuO}} = 68,97\% \)