Hòa tan hoàn toàn hỗn hợp H gồm Mg (5a mol) và  Fe3O4 (a  mol) trong  dung  dịch chứa KNO3

Phương pháp: Bảo toàn nguyên tố

\({n_{NH_4^ + }} = {n_{KN{O_3}}} - {n_{NO}} = \left( {x - 0,06} \right)\)

\({m_{muoi}} = {m_H} + {m_{{K^ + }}} + {m_{NH_4^ + }} + {m_{C{l^ - }}} - {m_{O\left( {trongH} \right)}} = 39x + 18\left( {x - 0,06} \right) + 35,5.0,725 - 64a\)

\(\Rightarrow \left\{ \begin{array}{l} {m_{muoi}} - {m_H} = 26,23\\ 10{n_{NH_4^ + }} + 2{n_{O\left( {trongH} \right)}} + 4{n_{NO}} + 2{n_{{H_2}}} = {n_{HCl}} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 57x - 64a = 1,5725\\ 10\left( {x - 0,06} \right) + 8a + 0,28 = 0,725 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 0,0725\\ a = 0,04 \end{array} \right.\)

\(\to {m_H} = 24,5a + 232a = 14,08\left( g \right) \Rightarrow {m_{muoi\,khan}} = 40,31\left( g \right)\)

\(\Rightarrow \% {m_{Fe}} = \frac{{0,04.3.56}}{{40,31}}.100\% = 16,67\% \)