\(a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ b)\\ V_{CO_2} = V_{CH_4} = 2,24(lít)\\ V_{O_2} = 2V_{CH_4} = 2,24.2 = 4,48(lít)\)
\(a.CH_4+2O_2\rightarrow CO_2+2H_2O\)
b.
\(n_{CH_4}=\dfrac{V}{22.4}=\dfrac{2.24}{22.4}=0.1mol\)
\(n_{O_2}=\dfrac{0,1.2}{1}=0.2mol\)
\(V_{O_2}=n.22,4=0,2.22,4=4.48l\)
\(n_{CO_2}=\dfrac{0,1.1}{1}=0.1mol\)
\(V_{CO_2}=n.22,4=0,1.22,4=2.24l\)