nCaCO3 = 30/100 = 0.3 mol
Ca(OH)2 + CO2 --> CaCO3 + H2O
__________0.3_______0.3
m dung dịch giảm = mCaCO3 - (mCO2 + mH2O ) = 6
<=> 30 - (mH2O + 0.3*44) = 6
<=> mH2O = 10.8 g
<=> nH2O = 0.6 mol
2CxHy + (4x + y)O2 -to-> 2xCO2 + yH2O
________________________2x______y
________________________0.3______0.6
<=> 0.6*2x = 0.3y
<=> x/y = 1/4
CTHH : CH4