nCaCO3=20/100=0.2mol 
n hh khí=3.36/22.4=0.15mol 
CH4+2O2-->CO2+2H2O 
amol---------->amol 
2C2H2+5O2-->4CO2+2H2O 
bmol------------->2bmol 
CO2+Ca(OH)2-->CaCO3+H2O 
0.2mol<-------------0.2mol 
nCO2=a+2b=0.2 (1) 
Ta có:n hh khí=a+b=0.15 (2) 
Từ (1)(2)=>a=0.1mol, b=0.05mol
%V=%n 
%VCH4=%nC=0.1.100/0.15=66.67%
=>%VC2H2=100-66.67=33.33%