Đáp án:
\(C{\% _{CuC{l_2}}} = 13,36\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Cu{(OH)_2} + 2HCl\xrightarrow{{}}CuC{l_2} + 2{H_2}O\)
Ta có:
\({n_{Cu{{(OH)}_2}}} = \frac{{19,6}}{{64 + 34}} = 0,2{\text{ mol;}}{{\text{m}}_{HCl}} = 182,5.10\% = 18,25{\text{gam}} \to {n_{HCl}} = \frac{{18,25}}{{36,5}} = 0,5{\text{ mol > 2}}{{\text{n}}_{Cu{{(OH)}_2}}}\) do vậy HCl dư
\( \to {n_{HCl{\text{dư}}}} = {n_{HCl}} - 2{n_{Cu{{(OH)}_2}}} = 0,5 - 0,2.2 = 0,1{\text{ mol}} \to {{\text{m}}_{HCl dư}} = 0,1.36,5 = 3,65{\text{ gam}}\)
BTKL:
\({m_{dd\;{\text{sau phản ứng}}}} = {m_{Cu{{(OH)}_2}}} + {m_{dd{\text{ HCL}}}} = 19,6 + 182,5 = 202,1{\text{ gam}}\)
\({n_{CuC{l_2}}} = {n_{Cu{{(OH)}_2}}} = 0,2{\text{ mol}} \to {{\text{m}}_{CuC{l_2}}} = 0,2.(64 + 35,5.2) = 27{\text{gam}}\)
\( \to C{\% _{CuC{l_2}}} = \frac{{27}}{{202,1}} = 13,36\% \)
