$1)C_\overline{n}H_{2\overline{n}+2}:x mol\Rightarrow n_{CO_2}=n_{CaCO_3}=\overline{n}x mol\Rightarrow \begin{cases}x=\frac{1,53}{14\overline{n}+2} \\ \frac{10,5}{100}=\overline{n}x\end{cases}\Rightarrow x=0,03 mol$$2)x=0,03 mol\Rightarrow \overline{n}=3,5\Rightarrow (C_4H_{10},CH_4) ; (C_4H_{10},C_2H_6) ; (C_4H_{10},C_3H_8)$