Giả sư: nCH4 =x(mol)

nC2H4 =y(mol)

-> 22,4x+22,4y=3,36(l)(*)

pt: CH4 + 2O2-to-> CO2 + 2H2O(1)

C2H4 + 3O2 -to-> 2CO2 + 2H2O(2)

CO2 + Ca(OH)2 -> CaCO3 +H2O

nCaCO3 =20:100=0,2(mol)

từ pt (1): nCO2=nCH4=x(mol)

pt (2): nCO2=2nC2H4=2y(mol)

-> x+2y=0,2(mol)(**)

từ (*) và (**)-> x=0,1(mol); y=0,05(mol)

-> VCH4=0,1.22,4=2,24(l)

-> %CH4=2,24:3,36.100%=66,67%

-> %C2H4= 100%-66,67%=33,33%

=>nCa(OH)2=0,1+0,05.2=0,2 mol

=>Cm=0,2\0,2=1M