Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
2,
Bảo toàn electron, ta có:
\(\begin{array}{l}
Al \to A{l^{3 + }} + 3e\\
{N^{5 + }} + 1e \to {N^{4 + }}\\
\to {n_{N{O_2}}} = 3{n_{Al}}\\
\to {n_{N{O_2}}} = 3 \times \dfrac{{5,4}}{{27}} = 0,6mol\\
\to {V_{N{O_2}}} = 13,44mol
\end{array}\)
3,
\(\begin{array}{l}
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
{n_{Cu}} = 0,25mol\\
\to {n_{NO}} = \dfrac{2}{3}{n_{Cu}} = 0,167mol\\
\to {V_{NO}} = 3,733l\\
\to {n_{HN{O_3}}} = \dfrac{8}{3}{n_{Cu}} = \dfrac{2}{3}mol\\
\to C{M_{HN{O_3}}} = \dfrac{{\dfrac{2}{3}}}{{0,25}} = 2,67M
\end{array}\)
4,
Khí hóa nâu trong không khí là NO
\(\begin{array}{l}
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + {H_2}O\\
{n_{Fe}} = 0,06mol\\
\to {n_{NO}} = {n_{Fe}} = 0,06mol\\
\to {V_{NO}} = 1,344l\\
\to {n_{HN{O_3}}} = 4{n_{Fe}} = 0,24mol\\
\to {m_{HN{O_3}}} = 15,12g\\
\to {m_{HN{O_3}}}{\rm{dd}} = \dfrac{{15,12 \times 100}}{{20}} = 75,6g\\
\to {V_{HN{O_3}}}{\rm{dd}} = \dfrac{{75,6}}{{1,4}} = 54c{m^3}
\end{array}\)
5,
\(\begin{array}{l}
Zn + 4HN{O_3} \to Zn{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
{n_{N{O_2}}} = 0,24mol\\
\to {n_{Zn}} = \dfrac{1}{2}{n_{N{O_2}}} = 0,12mol\\
\to m = {m_{Zn}} = 7,8g\\
\to {n_{HN{O_3}}} = 2{n_{N{O_2}}} = 0,48mol\\
\to {V_{HN{O_3}}} = \dfrac{{0,48}}{1} = 0,48l
\end{array}\)
