a,

$propan:C_{3}H_{8}$

$propin:C_{3}H_{4}$

$etilen:C_{2}H_{4}$

$m_{\text{mỗi phần}}=\frac{20,1}{3}=6,7$

⇒$44nC_{3}H_{8}+40nC_{3}H_{4}+28nC_{2}H_{4}=6,7 (1)$

P1:

$C_{3}H_{4}+2Br_{2} \to C_{3}H_{4}Br_{4}$

$C_{2}H_{4}+Br_{2} \to C_{2}H_{4}Br_{2}$

$nBr_{2}=0,5.0,39=0,195$

Theo  pt ta có:

⇒$2nC_{3}H_{4}+nC_{2}H_{4}=0,195(2)$

P2:

BTNT "C"

$nC_{3}H_{3}Ag=nC_{3}H_{4}=\frac{8,82}{147}=0,06(3)$

(1)(2)(3)⇒$nC_{3}H_{8}=0,05 ;nC_{3}H_{4}=0,06 ;nC_{2}H_{4}=0,075$

$\%mC_{3}H_{8}=\frac{0,05.44}{6,7}.100=32,84\%$

$\%mC_{3}H_{4}=\frac{0,06.40}{6,7}.100=35,82\%$

$\%mC_{2}H_{4}=100-32,84-35,82=31,34\%$

b,

BTNT "C"

$3nC_{3}H_{8}+3nC_{3}H_{4}+2nC_{2}H_{4}=nCO_{2}$

⇒$nCO_{2}=0,05.3+0,06.3+0,075.2=0,48$

$nBa(OH)_{2}=0,42.1=0,42$

$T=\frac{2nBa(OH)_{2}}{nCO_{2}}=\frac{2.0,42}{0,48}=1,75$

⇒Sinh ra 2 muối là $Ba(HCO_{3})_{2}$ và $BaCO_{3}$

Bảo toàn nguyên tố "Ba" và "C"

$nBa(HCO_{3})_{2}+nBaCO_{3}=nBa(OH)_{2}=0,42$

$2nBa(HCO_{3})_{2}+nBaCO_{3}=nCO_{2}=0,48$

⇒$\left \{ {{nBa(HCO_{3})_{2}=0,06} \atop {nBaCO_{3}=0,36}} \right.$ 

$mBaCO_{3}=0,36.197=70,92g$