Đáp án:
\(\begin{array}{l}
{m_{Cu{{(OH)}_2}}} = 19,6g\\
C{M_{NaOH}} = 0,8M\\
{m_{Cu{{(N{O_3})}_2}}} = 37,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
CuC{l_2} + 2NaOH \to 2NaCl + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O
\end{array}\)
\(\begin{array}{l}
b,\\
{n_{CuC{l_2}}} = 0,2mol\\
\to {n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,2mol\\
\to {m_{Cu{{(OH)}_2}}} = 19,6g\\
\to {n_{NaOH}} = 2{n_{CuC{l_2}}} = 0,4mol\\
\to C{M_{NaOH}} = \dfrac{{0,4}}{{0,5}} = 0,8M
\end{array}\)
\(\begin{array}{l}
c,\\
Cu{(OH)_2} + 2HN{O_3} \to Cu{(N{O_3})_2} + 2{H_2}O\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
\to {m_{Cu{{(N{O_3})}_2}}} = 37,6g
\end{array}\)