a,

Đặt CTTQ ankan: $C_nH_{2n+2}$

Sản phẩm thế clo: $C_nH_{2n+2-x}Cl_x$

$\%H=4,004\%$

$\Rightarrow \dfrac{2n+2-x}{14n+2-x+35,5x}=4,004\%$

$\Leftrightarrow 2n+2-x=4,004\%(14n+2+34,5x)$

$\Leftrightarrow 1,44n+1,92=2,38x$

$\Rightarrow n=x=2$

Vậy hợp chất hữu cơ là $C_2H_4Cl_2$

Ankan ban đầu là $C_2H_6$

PTHH:

$C_2H_6+2Cl_2\buildrel{{as}}\over\to 2HCl+ C_2H_4Cl_2$

b,

$n_{O_2}=\dfrac{7,84}{22,4}=0,35(mol)$

$C_2H_6+3,5O_2\buildrel{{t^o}}\over\to 2CO_2+3H_2O$

$\Rightarrow n_{C_2H_6}=\dfrac{0,35}{3,5}=0,1(mol)$

$PV=nRT\Rightarrow V=\dfrac{nRT}{P}$

$V_{C_2H_6}=\dfrac{0,1.0,082(12+273)}{1,25}=1,8696l$

$n_{CO_2}=0,35.2=0,7(mol)$

$n_{H_2O}=0,35.3=1,05(mol)$

$CO_2+Ba(OH)_2\to BaCO_3+H_2O$

$\Rightarrow n_{BaCO_3}=0,7(mol)$

$\Delta m_{dd}=m_{CO_2}+m_{H_2O}-m_{BaCO_3}$

$=0,7.44+1,05.18-0,7.197=-88,2g$

$\to dd$ giảm $88,2g$