Đáp án:
\(\begin{array}{l}
b){m_{Cu{{(OH)}_2}}} = 19,6g\\
c)\\
{m_{NaOH(dư)}} = 4g\\
{m_{NaCl}} = 23,4g\\
d)\\
C{\% _{NaOH(dư)}} = 14,6\% \\
C{\% _{NaCl}} = 85,4\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} \to CuO + {H_2}O\\
b)\\
{n_{NaOH}} = 0,5mol\\
\to {n_{NaOH}} > {n_{CuC{l_2}}} \to {n_{NaOH}}dư\\
{n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,2mol\\
\to {m_{Cu{{(OH)}_2}}} = 19,6g\\
c)\\
{n_{NaOH(pt)}} = {n_{NaCl}} = 2{n_{CuC{l_2}}} = 0,4mol\\
\to {n_{NaOH(dư)}} = 0,1mol\\
\to {m_{NaOH(dư)}} = 4g\\
\to {m_{NaCl}} = 23,4g\\
d)\\
{m_{{\rm{dd}}thuđược}} = {m_{CuC{l_2}}} + {m_{NaOH}} - {m_{Cu{{(OH)}_2}}} = 27,4g\\
\to C{\% _{NaOH(dư)}} = \dfrac{4}{{27,4}} \times 100\% = 14,6\% \\
\to C{\% _{NaCl}} = \dfrac{{23,4}}{{27,4}} \times 100\% = 85,4\% \\
\end{array}\)
