Hỗn hợp X gồm FeS, FeS2, Fex...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(29,2(g)\left\{ \begin{array}{l}Fe\\S\\{O_2}\end{array} \right\} + HN{O_3}:1,65 \to \left\{ \begin{array}{l}ddY\underbrace {\left\{ \begin{array}{l}HN{O_3}:0,03\\S{O_4}^{2 - }:a\\F{e^{3 + }}:b\\N{O_3}^ - du:c\end{array} \right\}}_{77,98(g)muoi} \overset{+Ba(OH)_{2du}}{\rightarrow} \underbrace {\left\{ \begin{array}{l}{\rm{BaS}}{O_4}:a\\F{e_2}{O_3}:0,5b\end{array} \right\}}_{83,92(g)}\\38,7(g)\left\{ \begin{array}{l}NO\\N{O_2}\end{array} \right\}\\\\{H_2}O\end{array} \right.\)

\[{n_{HN{O_3}pu}} = x(mol) \to {n_{{H_2}O}} = 0,5x(mol)\]

\(\xrightarrow{{BTKL}}{m_X} + {m_{HN{O_3}}} = {m_{muoi}} + {m_{khi}} + {m_{{H_2}O}}\)

\[\begin{array}{l} \to x = 1,62(mol)\\ \to {n_{HN{O_3}du}} = 0,03(mol)\end{array}\]

\(\left\{ \begin{array}{l} \overset{BTDT}{\rightarrow} 3b - 2a - c = 0\\{m_Y} = 96a + 56b + 62c = 77,98\\233a + 0,5b.160 = {m_{chat\,ran}} = 83,92\end{array} \right. \to \left\{ \begin{array}{l}a = 0,24\\b = 0,35\\c = 0,57\end{array} \right.\)

\[\begin{array}{l}3Cu\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,8{H^ + } + 2N{O_3}^ -  \to 3C{u^{2 + }} + 2NO + 4{H_2}O\\0,01125 \leftarrow 0,03\\Cu\,\,\,\,\,\,\,\, + \,\,\,\,\,2F{e^{3 + }} \to C{u^{2 + }} + 2F{e^{2 + }}\\0,175 \leftarrow 0,35\\ \to {m_{Cu}} = (0,01125 + 0,175).64 = 11,92(g)\end{array}\]

Đáp án C