Đáp án:
\(\% {m_{F{e_2}{O_3}}} = \dfrac{{16}}{{20}} \times 100\% = 80\% \)
Giải thích các bước giải:
\(\begin{array}{l}
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
{n_{Fe}} = \dfrac{{12 \times {{10}^{22}}}}{{6,023 \times {{10}^{23}}}} = 0,2mol\\
{n_{C{O_2}}} = 0,3mol\\
{n_{CO}} = 0,3mol\\
\to {n_{Fe}} = \dfrac{2}{3}{n_{CO}} = \dfrac{2}{3}{n_{C{O_2}}} = 0,2mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe}} = 0,1mol\\
\to {m_{F{e_2}{O_3}}} = 16g\\
\to \% {m_{F{e_2}{O_3}}} = \dfrac{{16}}{{20}} \times 100\% = 80\%
\end{array}\)
