Phản ứng xảy ra:
\(C + {O_2}\xrightarrow{{}}C{O_2}\)
Ta có: \({m_C} = 19.96\% = 18,24{\text{ kg}} \to {{\text{n}}_C} = \frac{{18,24}}{{12}} = 1,52{\text{ kmol}}\)
\({n_{kk}} = \frac{{2,24}}{{22,4}} = 0,1k{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{1}{5}{n_{kk}} = 0,02kmol < {n_C}\)
Vậy C dư.
(\(1kg = 1000{\text{ gam; 1 }}{{\text{m}}^3} = 1000\;{\text{lít}}\) nên đưa về cùng hệ)