Đốt cháy hidrocacbon:
$C_xH_y+\dfrac{4x+y}{2}O_2\to 2xCO_2+yH_2O$
$\%m_{CO_2}=61,972\%$
$\Rightarrow \dfrac{2x.44.100}{2x.44+y.18}=61,972$
$\Leftrightarrow 18y=54x$
$\Leftrightarrow \dfrac{x}{y}=\dfrac{1}{3}$
$\Rightarrow$ CTĐGN $(CH_3)_n$
$n=2\Rightarrow C_2H_6$
Vậy CTPT A là $C_2H_6$
$n_{Ca(OH)_2}=0,1(mol)$
$n_{CO_2}=0,08(mol)<0,1$
- TH1: dư $OH^-$
$\Rightarrow 2n_{C_2H_6}=n_{CO_2}=n_{CaCO_3}=0,08(mol)$
$\Rightarrow n_{C_2H_6}=0,04(mol)$
$V=0,04.22,4=0,896l$
- TH2: 2 muối
Bảo toàn Ca: $n_{Ca(HCO_3)_2}=0,1-0,08=0,02(mol)$
Bảo toàn C:
$2n_{C_2H_6}=n_{CO_2}=n_{CaCO_3}+2n_{Ca(HCO_3)_2}=0,12(mol)$
$\Rightarrow n_{C_2H_6}=0,06(mol)$
$V=0,06.22,4=1,344l$
