Hòa tan hoàn toàn 21,5 gam hỗn hợp X gồm Al, Zn, F...

C

- Hướng dẫn giải

- Khi cho 21,5 gam X tác dụng với 0,43 mol H₂SO₄ thì:

\({n_{{H_2}O}} = \frac{{{m_X} + 98{n_{{H_2}S{O_4}}} - 30{n_{NO}} - 2{n_{{H_2}}} - {m_Z}}}{{18}} = 0,26{\rm{ mol}}\)

\({n_{NH_4^ + }} = \frac{{2{n_{{H_2}S{O_4}}} - 2{n_{{H_2}O}} - 2{n_{{H_2}}}}}{4} = 0,02{\rm{ mol}} \Rightarrow {{\rm{n}}_{Cu{{\left( {N{O_3}} \right)}_2}}} = \frac{{{n_{NH_4^ + }} + {n_{NO}}}}{2} = 0,04{\rm{ mol}}\)

- Ta có  \({n_{O\left( {trong{\rm{ X}}} \right)}} = {n_{FeO}} = \frac{{2{n_{{H_2}S{O_4}}} - 10{n_{NH_4^ + }} - 4{n_{NO}} - 2{n_{{H_2}}}}}{2} = 0,08{\rm{ mol}}\)

- Xét hỗn hợp X ta có:

\(\left\{ \begin{array}{l}
3{n_{Al}} + 2{n_{Zn}} = 3{n_{NO}} + 2{n_{{H_2}}} + 8{n_{NH_4^ + }} = 0,6\\
27{n_{Al}} + 65{n_{Zn}} = {m_X} - 72{n_{FeO}} - 188{n_{Cu{{\left( {N{O_3}} \right)}_2}}} = 8,22
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{n_{Al}} = 0,16{\rm{ mol}}\\
{{\rm{n}}_{Zn}} = 0,06{\rm{ mol}}
\end{array} \right.\)

\( \Rightarrow \% {m_{Al}} = \frac{{27.0,16}}{{21,5}}.100 = 20,09\)