Đáp án:

\({m_{Cu{{(OH)}_2}}} = 9,8{\text{ gam}}\)

Giải thích các bước giải:

 Ta có: 

\({m_{CuC{l_2}}} = 6,75\% .200 = 13,5{\text{ gam}}\)

\( \to {n_{CuC{l_2}}} = \frac{{13,5}}{{64 + 35,5.2}} = 0,1{\text{ mol}}\)

Phản ứng xảy ra:

\(CuC{l_2} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + 2NaCl\)

Theo phản ứng:

\({n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,1{\text{ mol}}\)

\( \to {m_{Cu{{(OH)}_2}}} = 0,1.(64 + 17.2) = 9,8{\text{ gam}}\)

BTKL:

\({m_{dd{\text{sau phản ứng}}}} = {m_{dd{\text{ CuC}}{{\text{l}}_2}}} + {m_{dd{\text{NaOH}}}} - {m_{Cu{{(OH)}_2}}}\)

\( = 200 + 109,8 - 9,8 = 300{\text{ gam}}\)

\({n_{NaCl}} = 2{n_{CuC{l_2}}} = 0,2{\text{ mol}}\)

\( \to {m_{NaCl}} = 0,2.58,5 = 11,7{\text{ gam}}\)

\( \to C{\% _{NaCl}} = \frac{{11,7}}{{300}} = 3,9\% \)

Nung kết tủa:

\(Cu{(OH)_2}\xrightarrow{{{t^o}}}CuO + {H_2}O\)

\({n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,1{\text{ mol}}\)

\( \to {m_{CuO}} = 0,1.80 = 8{\text{ gam}}\)