Đáp án:

\({m_{CaC{O_3}}} = 10{\text{gam;}}{{\text{m}}_{CaO}} = 2,8{\text{ gam}}\)

\({C_{M{\text{ HCl}}}} = 2,5M\)

Giải thích các bước giải:

\(CaO + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O\)

\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)

\(Ca{(OH)_2} + 2HCl\xrightarrow{{}}CaC{l_2} + 2{H_2}O\)

Ta có: 

\({n_{C{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{CaC{O_3}}} \to {m_{CaC{O_3}}} = 0,1.100 = 10{\text{gam}} \to {{\text{m}}_{CaO}} = 2,8{\text{ gam}} \to {{\text{n}}_{CaO}} = \frac{{2,8}}{{40 + 16}} = 0,05{\text{ mol}}\)

Ta có: \({n_{Ca{{(OH)}_2}}} = 0,05.1 = 0,05{\text{ mol}} \to {{\text{n}}_{HCl{\text{ dư}}}} = 0,05.2 = 0,1{\text{ mol}}\)

\( \to {n_{HCl}} = 2{n_{CaC{O_3}}} + 2{n_{CaO}} + 2{n_{Ca{{(OH)}_2}}} = 2.0,1 + 2.0,05 + 2.0,05 = 0,4{\text{ mol}}\)

\({V_{dd{\text{ HCl}}}} = \frac{{200}}{{1,25}} = 160ml \to {C_{M{\text{ HCl}}}} = \frac{{0,4}}{{0,16}} = 2,5M\)

\({n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} + {n_{Ca{{(OH)}_2}}} = 0,1 + 0,05 + 0,05 = 0,2{\text{ mol}} \to {{\text{m}}_{CaC{l_2}}} = 0,2.111 = 22,2{\text{ gam}}\)