nhh=6,72\22,4=0,3 mol
=>nCh4=0,2 mol
nC2h4=0,1 mol
CH4+2O2-->CO2+2H2O
0,2-------------0,2 mol
C2H4+3O2-->2CO2+2H2O
0,1---------------0,05 mol
==>VCO2=(0,2+0,05).22,4=5,6 l
\(n_{hh}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{CH4}=0,2\left(mol\right)\)
\(CH_2+2O_2\rightarrow CO_2+2H_2O\)
\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)
\(n_{CO2}=0,2+0,1.2=0,4\left(mol\right)\)
\(\Rightarrow V_{CO2}=0,4.22,4=8,96\left(l\right)\)
