3a=11b =>b=3a/11
CO2 a gam=> mC=3a/11 =b (g)=>nC=b/12 mol
H2O b gam=> mH=b/9 (g)=>nH=b/9 mol
7x=3(a+b)=3(11b/3+b)=14b =>x=2b (g)
mC+mH=10b/9<2b ==>trongA có O (2b-10b/9)=8b/9 (g) =>nO=b/18 mol
C:H:O=(b/12):(b/9):(b/18)=3:4:2