1)

Phản ứng xảy ra:

\(C + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)

Ta có:

\({m_C} = 100.(100\%  - 15\% ) = 85{\text{ kg}}\)

\( \to {n_C} = \frac{{85}}{{12}} = {n_{C{O_2}{\text{ lt}}}}\)

\( \to {n_{C{O_2}}} = \frac{{85}}{{12}}.80\%  = 5,67{\text{ kmol}}\)

\( \to {m_{C{O_2}}} = 5,67.44 = 249,48{\text{ kg}}\)

2)

Phản ứng xảy ra:

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

Ta có:

\({m_{CaC{O_3}}} = 20.(100\%  - 25\% ) = 15{\text{ kg}}\)

\( \to {n_{CaC{O_3}}} = \frac{{15}}{{100}} = 0,15{\text{ kmol = }}{{\text{n}}_{C{O_2}{\text{ lt}}}}\)

\( \to {n_{C{O_2}}} = 0,15.75\%  = 0,1125{\text{ kmol}}\)

BTKL:

\({m_{CaC{O_3}}} = {m_{rắn}} + {m_{C{O_2}}}\)

\( \to {m_{rắn}} = 20 - 0,1125.44 = 15,05{\text{ kg}}\)