$\text{bài giải}$
$\text{nCO2}=$ $\dfrac{1,792}{22,4}=0,08 mol$
$\text{nH2O}=$ $\dfrac{1,62}{18}=0,09 mol$
$\text{Đốt cháy ankan: n ankan=nH2O-nCO2=0,09-0,08=0,01 mol}$
$\text{2n ankan=n anken=0,02 mol}$
$\text{Khối lượng bình tăng = khối lượng anken=0,84 gam}$
$\overline{M}anken =$$\dfrac{m}{n}=$$\dfrac{0,84}{0,02}=42$
$\overline{M}=14n=42$ $\Rightarrow$ $\text{n=3}$ $\Rightarrow$$\text{C3H6}$
$\text{2}$$C_{3}$$H_{6}+9$ $O_{2}$$\rightarrow$ $6CO_{2}+$ $6H_{2}O$
$\text{0,02 0,06 0,06}$
$\text{}$$n_{CO2}=$ $n_{CO2(A)}+$ $n_{CO2(B)}$
$\text{0,08=}$$n_{CO2(A)}+0,06$ $\Rightarrow$ $n_{CO2(A)}=0,02 mol$
$C_{n}$ $H_{2n+2}+$ $\dfrac{3n+1}{2}$$O_{2}$$\rightarrow$ $n{}$ $CO_{2}+$ $(n+1)H_{2}O$
$\text{0,01 0,02}$
$\dfrac{1}{0,01}=$$\dfrac{n}{0,02}$$\Rightarrow$$\text{n=2}$$\Rightarrow$$C_{2}$$H_{6}$
$m_{a}=$$m_{C3H6}+$$m_{C2H6}=0,02.42+0,01.30=1,14g$
