Số mol CO = \(\dfrac{{7,84}}{{22,4}} = 0,35(mol)\) .
\(CuO\,\,\,\,\, + \,\,\,\,\,\,CO\buildrel {{t^o}} \over\longrightarrow Cu\,\,\,\,\, + \,\,\,\,C{O_2}\)
x mol x mol x mol x mol
\(F{e_2}{O_3} + 3CO\buildrel {{t^o}} \over\longrightarrow 2Fe + 3C{O_2}\)
y mol 3y mol 2y mol 3y mol
Ta có phương trình:
\(\left\{ \matrix{x + 3y = 0,35 \hfill \cr 80x + 160y = 20 \hfill \cr} \right. \Rightarrow \left\{ \matrix{x = 0,05 \hfill \cr y = 0,1 \hfill \cr} \right.\)
\(\% {m_{CuO}} = \dfrac{{80 \times 0,05 \times 100\%}}{{20}} = 20\% ;\% {m_{F{e_2}{O_3}}} = 100\% - 20\% = 80\% \)
