Đáp án:

\({m_{dd{\text{ HCl}}}}  = 39,46{\text{ gam}}\)

\( {m_{Ca}}  = 4{\text{ gam;}}{{\text{m}}_{CaO}} = 5,6{\text{ gam}}\)

\(\% {m_{Ca}} = 41,67\% ; \% {m_{CaO}} = 58,33\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

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

Ta có:

\({n_{CaC{l_2}}} = \frac{{22,2}}{{40 + 35,5.2}} = 0,2{\text{ mol = }}{{\text{n}}_{Ca}} + {n_{CaO}}\)

Ta có:

\({n_{HCl}} = 2{n_{CaC{l_2}}} = 0,4{\text{ mol}}\)

\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)

\( \to {m_{dd{\text{ HCl}}}} = \frac{{14,6}}{{37\% }} = 39,46{\text{ gam}}\)

Ta có:

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Ca}}\)

\({n_{CaO}} = {n_{CaC{l_2}}} - {n_{Ca}} = 0,2 - 0,1 = 0,1{\text{ mol}}\)

\( \to {m_{Ca}} = 0,1.40 = 4{\text{ gam;}}{{\text{m}}_{CaO}} = 0,1.56 = 5,6{\text{ gam}}\)

\( \to \% {m_{Ca}} = \frac{4}{{4 + 5,6}} = 41,67\%  \to \% {m_{CaO}} = 58,33\% \)