Đáp án:

b. \(C{\% _{HCl}} = \dfrac{{14,6}}{{73}} \times 100\%  = 20\% \)

c. \({m_{CuO}} = 0,015g\)

Giải thích các bước giải:

\(\begin{array}{l}
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuO}} = 0,2mol\\
 \to {n_{HCl}} = 2{n_{CuO}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to C{\% _{HCl}} = \dfrac{{14,6}}{{73}} \times 100\%  = 20\% \\
2NaOH + CuC{l_2} \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{m_{NaOH}} = D \times V = 1,25 \times 0,3 = 0,375g\\
 \to {n_{NaOH}} = \dfrac{{0,375 \times 4}}{{100 \times 40}} = 0,000375mol\\
{n_{CuC{l_2}}} = {n_{CuO}} = 0,2mol\\
 \to \dfrac{{{n_{NaOH}}}}{2} < {n_{CuC{l_2}}}
\end{array}\)

Suy ra \(CuC{l_2}\) dư

\(\begin{array}{l}
 \to {n_{Cu{{(OH)}_2}}} = {n_{CuO}} = \dfrac{1}{2}{n_{NaOH}} = 0,0001875mol\\
 \to {m_{CuO}} = 0,015g
\end{array}\)