Đáp án:
\( {m_{dd{\text{ Ca(OH}}{{\text{)}}_2}}} = 148{\text{ gam}}\)
\( C{\% _{CaC{l_2}}} = 6,023\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Ca{(OH)_2} + 2HCl\xrightarrow{{}}CaC{l_2} + 2{H_2}O\)
Ta có:
\({m_{dd\;{\text{HCl}}}} = 200.1,1 = 220{\text{ gam}}\)
\({n_{HCl}} = 0,2.2 = 0,4{\text{ mol}}\)
\( \to {n_{Ca{{(OH)}_2}}} = \frac{1}{2}{n_{HCl}} = 0,2{\text{ mol}}\)
\( \to {m_{Ca{{(OH)}_2}}} = 0,2.(40 + 17.2) = 14,8{\text{ gam}}\)
\( \to {m_{dd{\text{ Ca(OH}}{{\text{)}}_2}}} = \frac{{14,8}}{{10\% }} = 148{\text{ gam}}\)
\( \to {m_{dd{\text{ muối}}}} = {m_{dd{\text{ Ca(OH}}{{\text{)}}_2}}} + {m_{dd\;{\text{HCl}}}} = 220 + 148 = 368{\text{ gam}}\)
\({n_{CaC{l_2}}} = {n_{Ca{{(OH)}_2}}} = 0,2{\text{ mol}}\)
\( \to {m_{CaC{l_2}}} = 0,2.(40 + 35,5.2) = 22,2{\text{ gam}}\)
\( \to C{\% _{CaC{l_2}}} = \frac{{22,2}}{{368}}.100\% = 6,023\% \)