Câu 8:
a.
$m_{H_2O}=12,6g$
$\to n_{H_2O}=\dfrac{12,6}{18}=0,7(mol)$
$m_C+m_H=6,8g$
$\to n_{CO_2}=n_C=\dfrac{6,8-0,7.2}{12}=0,45(mol)$
Hỗn hợp $X$ gồm 2 ankan nên $n_X=n_{H_2O}-n_{CO_2}=0,25(mol)$
$n_A: n_{CH_4}=2:3$
$\to n_A=\dfrac{0,25}{2+3}.2=0,1(mol)$
$\to n_{CH_4}=0,25-0,1=0,15(mol)$
Đặt CTTQ ankan $A$ là $C_nH_{2n+2}$
Bảo toàn $C$: $0,1n+0,15=0,45$
$\to n=3$
Vậy CTPT ankan $A$ là $C_3H_8$
b,
$m_{CH_4}=0,15.16=2,4g$
$m_{C_3H_8}=0,1.44=4,4g$
c,
$n_{Ba(OH)_2}=1,2.0,25=0,3(mol)$
$\dfrac{2n_{Ba(OH)_2}}{n_{CO_2}}=1,33$
$\to$ tạo muối $BaCO_3$ (x mol), $Ba(HCO_3)_2$ (y mol)
Bảo toàn $Ba$: $x+y=0,3$
Bảo toàn $C$: $x+2y=0,45$
$\to x=y=0,15$
$\to m_{\text{muối}}=197.0,15+259.0,15=68,4g$
