CxHy + \(\left(x+\dfrac{y}{4}\right)\)O2 -to-> xCO2 + y/2H2O (1)
CO2 +Ca(OH)2 --> CaCO3 +H2O (2)
2CO2 +Ca(OH)2 --> Ca(HCO3)2 (3)
nCaCO3 =0,1(mol)
theo (2) : nCa(OH)2(2)=nCO2=nCaCO3=0,1(mol)
=> nCa(OH)2(3)=0,4.0,5-0,1=0,1(mol)
theo (3) : nCO2(3)=2nCa(OH)2(3)=0,2(mol)
=> \(\Sigma nCO2\)=0,3(mol)
=> mCO2=13,2(g)
=>mH2O=18,6-13,2=5,4(mol)
=>nH2O=0,3(mol)
nCxHy=0,1(mol)
theo (1) : nCxHy=1/x nCO2
=>\(\dfrac{1}{x}=\dfrac{0,1}{0,3}=>x=3\)
\(\dfrac{nCxHy}{nH2O}=\dfrac{2}{y}=\dfrac{0,1}{0,3}=>y=6\)
=> CxHy: C3H6