Đáp án:
b.
mCu=7,84g
mCuO=3,96g
c.
V=1,12l
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
b.\\
{n_{N{O_2}}} = 0,245mol\\
\to {n_{Cu}} = \dfrac{1}{2}{n_{N{O_2}}} = 0,1225mol\\
\to {m_{Cu}} = 7,84g\\
\to {m_{CuO}} = 3,96g\\
c.\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
{n_{HN{O_3}}} = 0,2mol\\
\to \dfrac{{{n_{Cu}}}}{3} > \dfrac{{{n_{HN{O_3}}}}}{8} \to {n_{Cu}}dư\\
\to {n_{NO}} = \dfrac{1}{4}{n_{HN{O_3}}} = 0,05mol\\
\to {V_{NO}} = 1,12l
\end{array}\)
