10)

Phản ứng xảy ra:

\(CaC{O_3}\xrightarrow{{{t^o}}}CaO + C{O_2}\)

Ta có: 

\({n_{C{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{CaC{O_3}{\text{ lt}}}} = {n_{CaO}} \to {m_{CaO}} = 0,1.(40 + 16) = 5,6{\text{ gam}}\)

\({n_{CaC{O_3}{\text{ thực tế}}}} = \frac{{{n_{CaC{O_3}lt}}}}{h} = \frac{{0,1}}{h} \to {m_{CaC{O_3}}} = \frac{{0,1}}{h}.100 = \frac{{10}}{h} \to a = \frac{{{m_{CaC{O_3}}}}}{{70\% }} = \frac{{\frac{{10}}{h}}}{{70\% }} = \frac{{100}}{{7h}}{\text{ gam}}\)

11)

Ta có: \({n_{C{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol; }}{{\text{n}}_{Ca{{(OH)}_2}}} = 0,2.0,5 = 0,1{\text{ mol}} \to \frac{{{n_{C{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \frac{{0,15}}{{0,1}} = 1,5\) do vậy tạo ra 2 muối.

Phản ứng xảy ra:

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

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

Vì 1,5 là trung bình của 1 và 2.

\( \to {n_{CaC{O_3}}} = {n_{Ca{{(HC{O_3})}_2}}} = \frac{{0,1}}{2} = 0,05{\text{ mol}} \to {{\text{m}}_{CaC{O_3}}} = 0,05.100 = 5{\text{ gam}}\)