Nung m gam bột sắt trong oxi, thu được 4,5 gam hỗn...

A

- Hướng dẫn giải

\(\left\{ \begin{gathered} {n_{Fe}}:x \hfill \\ {n_O}:y \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} \xrightarrow[{}]{{BTKL}}56x + 16y = 4,5 \hfill \\ \xrightarrow[{}]{{BT:e}}3x - 2y = 2{n_{S{O_2}}} = 0,1125 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} x = 0,0675{\text{ }}\left( {mol} \right) \hfill \\ y = 0,045{\text{ }}\left( {mol} \right) \hfill \\ \end{gathered} \right.\)

\({m_{Fe}} = 0,0675.56 = 3,78\left( {gam} \right)\)