Hỗn hợp X gồm Mg, Fe, Fe3O4...

D

- Hướng dẫn giải

\(m\left( {Mg,Fe,Cu} \right) = 0,8a\left( g \right);{m_O} = 0,2a\left( g \right)\). Đặt \({n_{{H_2}S{O_4}}} = 1,32x \Rightarrow {n_{NaN{O_3}}} = 0,8x\).

\(\left\{ {\left( {Mg,Fe,Cu} \right);O} \right\} + \left\{ {{H_2}S{O_4}:1,32x;NaN{O_3}:0,8x} \right\} \to NO:0,08 + {H_2}O\left\{ {\left( {M{g^{2 + }},F{e^{? + }},C{u^{2 + }}} \right);N{a^ + }:0,8x;NH_4^ + ;SO_4^{2 - }:1,32x;NO_3^ - } \right\}\) 

\(\left\{ {\left( {M{g^{2 + }},F{e^{? + }},C{u^{2 + }}} \right);N{a^ + }:0,8x;NH_4^ + ;SO_4^{2 - }:1,32x;NO_3^ - } \right\} + KOH:1,22 \to \left\{ {{K^ + }:1,22;N{a^ + }:0,8x;SO_4^{2 - }:1,32x;NO_3^ - } \right\}\) 

\({n_{{H^ + }}} = 2{n_O} + 10{n_{NH_4^ + }} + 4{n_{NO}} \Rightarrow {n_{NH_4^ + }} = \left( {0,264x - 0,0025a - 0,032} \right){\rm{ mol}}\).

Bảo toàn nguyên tố nitơ: \({n_{NO_3^ - /Y}} = {n_{NO}} + {n_{NH_4^ + }} + {n_{NO_3^ - /Z}} \Rightarrow {n_{NO_3^ - /Z}} = 0,536x + 0,0025a - 0,048\) 

Bảo toàn điện tích: \({n_{{K^ + }}} + {n_{N{a^ + }}} = 2{n_{SO_4^{2 - }}} + {n_{NO_3^ - }} \Rightarrow 1,22 + 0,8x = 3,176x + 0,0025a - 0,048\) 

\(b = m\left( {Mg,Fe,Cu} \right) + {m_{N{a^ + }}} + {m_{NH_4^ + }} + {m_{SO_4^{2 - }}} + {m_{NO_3^ - }} = 183,104x + 0,91a - 3,552\]) 

\(\Rightarrow 183a = 50 \times \left( {183,104x + 0,91a - 3,552} \right)\) 

Giải hệ có: \(x = 0,5{\rm{ mol;a = 32}}\left( g \right) \Rightarrow b = 117,12\left( g \right)\).