Đáp án:
\(\begin{array}{l}
a)\\
V = 500ml\\
b)\\
{C_\% }CuC{l_2} = 7,11\% \\
{C_\% }HCl = 3,08\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuC{l_2} + 2KOH \to Cu{(OH)_2} + 2KCl\\
{n_{KOH}} = 0,25 \times 2 = 0,5\,mol\\
{n_{CuC{l_2}}} = \dfrac{{0,5}}{2} = 0,25\,mol\\
V = {V_{{\rm{dd}}CuC{l_2}}} = \dfrac{{0,25}}{{0,5}} = 0,5l = 500ml\\
b)\\
Cu{(OH)_2} + 2HCl \to CuC{l_2} + 2{H_2}O\\
{n_{HCl}} = \dfrac{{450 \times 7,3\% }}{{36,5}} = 0,9\,mol\\
{n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,25\,mol\\
{n_{HCl}}\text{ dư} = 0,9 - 0,25 \times 2 = 0,4\,mol\\
{n_{CuC{l_2}}} = {n_{Cu{{(OH)}_2}}} = 0,25\,mol\\
m_{CuC{l_2}} =0,25 \times 135 = 33,75 g\\
m_{HCl} \text{ dư}= 0,4 \times 36,5 =14,6g
\end{array}\)
