Đáp án:
b) \({m_{CuO}} = 10{\text{ gam; }}{{\text{m}}_{F{e_2}{O_3}}} = 16{\text{ gam}}\)
c) \({{\text{V}}_{{H_2}}} = 9,52{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)
\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)
Gọi số mol Cu là x; Fe là y
\( \to 64x + 56y = 19,2{\text{ gam}}\)
\(\% {m_{Cu}} = \frac{{64x}}{{19,2}} = 41,67\% \to x = 0,125{\text{ mol}} \to {\text{y = 0}}{\text{,2 mol}}\)
\( \to {n_{CuO}} = {n_{Cu}} = 0,125{\text{ mol; }}{{\text{n}}_{F{e_2}{O_3}}} = \frac{1}{2}{n_{Fe}} = 0,1{\text{ mol}}\)
\( \to {m_{CuO}} = 0,125.(64 + 16) = 10{\text{ gam; }}{{\text{m}}_{F{e_2}{O_3}}} = 0,1.(56.2 + 16.3) = 16{\text{ gam}}\)
\({n_{{H_2}}} = {n_{CuO}} + 3{n_{F{e_2}{O_3}}} = 0,125 + 3.0,1 = 0,425{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,425.22,4 = 9,52{\text{ lít}}\)