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:

\(\begin{array}{l}
\left\{ \begin{array}{l}
56{n_{Fe}} + 27{n_{Al}} = 14,07\\
3{n_{Fe}} + 3{n_{Al}} = 1,08
\end{array} \right.\\
 \to {n_{Fe}} = 0,15 \to {n_{Al}} = 0,21\\
 \to {m_{Al}} = 5,67g
\end{array}\)