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"
$nBa(HCO_{3})_{2}+nBaCO_{3}=nBa(OH)_{2}=0,42$
$2nBa(HCO_{3})_{2}+nBaCO_{3}=nCO_{2}=0,48$
⇒$\left \{ {{nBa(HCO_{3})_{2}=0,06} \atop {nBaCO_{3}=0,36}} \right.$
$mBaCO_{3}=0,36.197=70,92g$
