$C_nH_{2n+2}~+~\frac{3n+1}{2}O_2-->nCO_2~+~(n+1)H_2O$

$CO_2~+~Ca(OH)_2-->CaCO_3~+~H_2O~(~1~)$

$2CO_2~+~Ca(OH)_2-->Ca(HCO_3)_2~(~2~)$

$Ca(HCO_3)_2~+~Ba(OH)_2-->BaCO_3~+~CaCO_3~+~2H_2O~(~3~)$

$n_{Ca(OH)_2}=0,05~.~1,6=0,08~(~mol~)$.

$Đặt~số~mol~Ca(OH)_2~ở~(~1~),~(~2~)~lần~lượt~là~a~và~b~(~mol~)$

$\left \{ {{a~+~b=0,08~(~mol~)} \atop {100a~+~397b=11,94~(~g~)}} \right.$ 

$-->a=0,06~(~mol~)~và~b=0,02~(~mol~)$

$Δ^mDD=0,56=m_{CO_2}~+~m_{H_2O}-m_{CaCO_3}$

                    $=0,1~.~44~+~18~.~n_{H_2O}~-~0,06.100$

$-->n_{H_2O}=0,12~(~mol~)$

$-->\frac{n}{2n+2}=\frac{0,1}{0,24}-->n=5$

$Vậy~CTPT~của~X~là~C_5H_{12}$