a,
nCO2= nBaCO3= 0,6 mol
\(\Rightarrow m_{CO2}=26,4\left(g\right)\)
m dd giảm= mBaCO3 - mCO2 - mH2O
\(\Rightarrow m_{H2O}=14,4\left(g\right)\)
\(\Rightarrow n_{H2O}=0,8\left(mol\right)\)
nCO2 < nH2O \(\Rightarrow\) Ankan CnH2n+2
\(n_A=0,2\left(mol\right)\)
Bảo toàn C:
\(0,6=0,2n\)
\(\Rightarrow n=3\)
Vậy A là C3H8
b,
CTCT là CH3-CH2-CH3