Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
\(\begin{array}{l}
6.\\
Fe + 6HN{O_3} \to Fe{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
FeO + 4HN{O_3} \to Fe{(N{O_3})_3} + N{O_2} + 2{H_2}O\\
{n_{Fe}} = 0,2mol\\
{n_{FeO}} = 0,1mol\\
\to {n_{N{O_2}}} = 3{n_{Fe}} + {n_{FeO}} = 0,7mol\\
\to {V_{N{O_2}}} = 15,68l
\end{array}\)
\(\begin{array}{l}
7.\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + 2{H_2}O\\
{n_{NO}} = 0,15mol
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{Cu}} = \dfrac{3}{2}{n_{NO}} = 0,225mol\\
\to {m_{Cu}} = 14,4g\\
\to {m_{CuO}} = 15 - 4,4 = 0,6g \to {n_{CuO}} = 0,0075mol
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{HN{O_3}}} = \dfrac{8}{3}{n_{Cu}} + 2{n_{CuO}} = 0,615mol\\
\to {m_{HN{O_3}}} = 38,7g\\
\to {m_{{\rm{dd}}HN{O_3}}} = \dfrac{{38,7}}{{10\% }} \times 100\% = 387,45g
\end{array}\)
\(\begin{array}{l}
8.\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
F{e_2}{O_3} + 6HN{O_3} \to 2Fe{(N{O_3})_3} + 3{H_2}O\\
{n_{NO}} = 0,075mol
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{Fe}} = {n_{NO}} = 0,075mol\\
\to {m_{Fe}} = 4,2g\\
\to {m_{F{e_2}{O_3}}} = 8g \to {n_{F{e_2}{O_3}}} = 0,05mol
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{HN{O_3}}} = 4{n_{Fe}} + 6{n_{F{e_2}{O_3}}} = 0,6mol\\
\to {m_{HN{O_3}}} = 37,8g\\
\to {m_{{\rm{dd}}HN{O_3}}} = \dfrac{{37,8}}{{10\% }} \times 100\% = 378g
\end{array}\)
