Đáp án:
22/
$nEtilen=\frac{4,48}{22,4}=0,2$
$C_{2}H_{4} +Br_{2} → C_{2}H_{4}Br_{2}$
0,2 0,2
$VddBr_{2}=\frac{0,2}{1}=0,2lit=200ml$
⇒B
23/
Anken $C_{n}H_{2n}$
nAnken=$\frac{3,36}{22,4}=0,15$
M Anken=$\frac{7,7}{0,15}$=51,33
⇒14n=51,33
⇔n=3,67
⇒Là $C_{3}H_{6}$ và $C_{4}H_{8}$
⇒B
24/
Gỉa tương tự bài 23
⇒$C_{3}H_{6}$ và $C_{4}H_{8}$
$nC_{3}H_{6}+ nC_{4}H_{8}=0,2$
$42nC_{3}H_{6}+56nC_{4}H_{8}=9,8$
⇒$nC_{3}H_{6}=nC_{4}H_{8}=0,1$
V%C3H6=n%C3H6=$\frac{0,1}{0,2}$.100=50%
⇒A
23/
nC=nCO2=$\frac{5,6}{22,4}$=0,25
nAnken=$\frac{1,12}{22,4}$=0,05$
nC:nAnken=0,25:0,05=5
⇒$C_{5}H_{10}$
⇒D
24/
$nBr_{2}=\frac{16}{160}=0,1$
$C_{n}H_{2n}+Br_{2} → C_{n}H_{2n}Br_{2}$
0,1 0,1
MAnken=$\frac{4,2}{0,1}$=42
⇒14n=42⇒n=3
⇒$C_{3}H_{6}$
CTCT: $CH_{2}=CH-CH_{3}$ : prop-1-en
$CH_{2}=CH-CH_{3} +H_{2} \buildrel{{Ni,t^o}}\over\longrightarrow CH_{3}-CH_{2}-CH_{3}$
$CH_{2}=CH-CH_{3} +HCl \buildrel{{t^o}}\over\longrightarrow CH_{3}-CHCl-CH_{3}$ (sản phẩm chính)
$CH_{2}=CH-CH_{3} +HCl \buildrel{{t^o}}\over\longrightarrow CH_{2}Cl-CH_{2}-CH_{3}$ (Sản phẩm phụ)
$CH_{2}=CH-CH_{3} \buildrel{{t^o,P,xt}}\over\longrightarrow -[CH_{2}-CH-CH_{3}]- (PE,polipropilen)$
