Đáp án:

 CT Ankan : $C_{\overline{n}}H_{2\overline{n}+2}$

$nH_{2}O=\frac{3,96}{18}=0,22$

$(\overline{n}+2)nH_{Ankan}=2nH_{2}O$

⇒$nH_{Ankan}=\frac{nH_{2}O}{(\overline{n}+1)}=\frac{0,22}{(\overline{n}+1)}$

$M Ankan=\frac{2,36}{0,22(\overline{n}+1)}$

⇔$12\overline{n} +2\overline{n}+2=\frac{2,36}{\frac{0,22}{\overline{n}+1}}$

⇔$\overline{n}=2,67$

⇒A : $C_{2}H_{6}$ và $CH_{3}-CH_{3}$

   B: $C_{3}H_{8}$ và $CH_{3}-CH_{2} -CH_{3}$

$n_{hh}=\frac{31,36}{22,4}=1,4$

$nC_{2}H_{4}=\frac{7,84}{28}=0,28$

$nC_{2}H_{6}+ nC_{3}H_{8}=1,4-0,28=1,12$(1)

$2nC_{2}H_{6}+ 3nC_{3}H_{8}=nCaCO_{3}+2nCa(HCO_{3})_{2}$

⇔$2nC_{2}H_{6}+ 3nC_{3}H_{8}=\frac{120}{100}+2.\frac{140,94}{162}$(2)

(1)(2)⇒$\left \{ {{nC_{2}H_{6}=0,42} \atop {nC_{3}H_{8}=0,7}} \right.$ 

$\%VC_{2}H_{6}=\%nC_{2}H_{2}=\frac{0,42}{1,4}.100=30\%$

$\%VC_{3}H_{8}=\%nC_{3}H_{8}=\frac{0,7}{1,4}.100=50\%$

$\%VC_{2}H_{4}=100-30-50=20\%$

$Mhh=\frac{0,7.44+0,42.30+7,84}{1,4}=36,6$

$d\frac{hh}{N_{2}}=\frac{36,6}{28}=1,307$