Đáp án:

\({C_{M{\text{ Ca(HC}}{{\text{O}}_3}{)_2}}} = 0,333M\)

\(C{\% _{Ca{{(HC{O_3})}_2}}} = 4,348\% \)

Giải thích các bước giải:

 Ta có: 

\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol;}}{{\text{n}}_{Ca{{(OH)}_2}}} = 0,15.1 = 0,15{\text{ mol}}\)

\( \to 1 < \frac{{{n_{C{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \frac{{0,2}}{{0,15}} < 2\)

Do vậy tạo ra 2 muối.

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

\(Ca{(OH)_2} + 2C{O_2}\xrightarrow{{}}Ca{(HC{O_3})_2}\)

Ta có: 

\({n_{Ca{{(HC{O_3})}_2}}} = {n_{C{O_2}}} - {n_{Ca{{(OH)}_2}}} = 0,05{\text{ mol}}\)

\( \to {n_{CaC{O_3}}} = 0,15 - 0,05 = 0,1{\text{mol}}\)

\( \to {C_{M{\text{ Ca(HC}}{{\text{O}}_3}{)_2}}} = \frac{{0,05}}{{0,15}} = 0,333M\)

\({m_{dd{\text{ Ca(OH}}{{\text{)}}_2}}} = 150.1,25 = 187,5{\text{ gam}}\)

BTKL: 

\({m_{dd\;{\text{X}}}} = {m_{C{O_2}}} + {m_{dd{\text{ Ca(OH}}{{\text{)}}_2}}} - {m_{CaC{O_3}}}\)

\( = 0,2.44 + 187,5 - 0,1.100 = 186,3{\text{ gam}}\)

\({m_{Ca{{(HC{O_3})}_2}}} = 0,05.(40 + 61.2) = 8,1{\text{ gam}}\)

\( \to C{\% _{Ca{{(HC{O_3})}_2}}} = \frac{{8,1}}{{186,3}} = 4,348\% \)