Đáp án:
39,C
40,B
Giải thích các bước giải:
39,
\(\begin{array}{l}
K + {H_2}O \to KOH + \dfrac{1}{2}{H_2}\\
{n_K} = \dfrac{m}{{39}}mol\\
\to {n_{KOH}} = {n_K} = \dfrac{m}{{39}}mol\\
\to {n_{{H_2}}} = \dfrac{1}{2} \times {n_K} = \dfrac{m}{{78}}mol\\
{m_{{\rm{dd}}}} = {m_K} + {m_{{H_2}O}} - {m_{{H_2}}} = m + 192,4 - (\dfrac{m}{{78}} \times 2) = (\dfrac{{38}}{{39}}m + 192,4)g\\
\to C{\% _{KOH}} = \dfrac{{\dfrac{m}{{39}} \times 56}}{{\dfrac{{38}}{{39}}m + 192,4}} \times 100 = 5,6\\
\to m = 7,9g\\
\to {n_{{H_2}}} = \dfrac{{7,9}}{{78}} = 0,1mol\\
\to {V_{{H_2}}} = 2,24l
\end{array}\)
40,
\({n_{NO}} + {n_{N{O_2}}} = 0,48mol\)
Áp dụng sơ đồ đường chéo ta có:
\(\dfrac{{{n_{NO}}}}{{{n_{N{O_2}}}}} = \dfrac{{46 - 9 \times 4}}{{9 \times 4 - 30}} = \dfrac{5}{3}\)
Giải hệ phương trình ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{n_{NO}} + {n_{N{O_2}}} = 0,48\\
\dfrac{{{n_{NO}}}}{{{n_{N{O_2}}}}} = \dfrac{5}{3}
\end{array} \right.\\
\to {n_{NO}} = 0,3 \to {n_{N{O_2}}} = 0,18
\end{array}\)
Bảo toàn electron ta có:
\(\begin{array}{l}
Fe \to F{e^{3 + }} + 3e\\
Al \to A{l^{3 + }} + 3e\\
{N^{5 + }} + 3e \to {N^{2 + }}\\
{N^{5 + }} + 1e \to {N^{4 + }}\\
\to 3{n_{Fe}} + 3{n_{Al}} = 3{n_{NO}} + {n_{N{O_2}}}\\
\to 3{n_{Fe}} + 3{n_{Al}} = 1,08
\end{array}\)
Mặt khác ta có:\(56{n_{Fe}} + 27{n_{Al}} = 14,07\)
Giải hệ phương trình ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
56{n_{Fe}} + 27{n_{Al}} = 14,07\\
3{n_{Fe}} + 3{n_{Al}} = 1,08
\end{array} \right.\\
\to {n_{Fe}} = 0,15 \to {n_{Al}} = 0,21\\
\to {m_{Al}} = 5,67g
\end{array}\)
