Question

Se iau 2 hidrocarburi A si B. Substanta A contine 6,96% H iar B are raportul de masa C:H=6:1. Cunoscand ca 1 g din substanta A ocupa 860 ml in c.n. iar 0,1 moli B aditioneaza 16 g Br si formeaza 20,2 g produs de aditie,sa se determine :
a) A si B si volumul de solutie de K2Cr2O7. 0,1M care in mediu acid oxideaza 8,4 grame B.

Answer 1

n=m/M=V/Vm => M=26g/mol

%C=93.04

nC=%C*M/AC*100 = 93,04*26/1200 = 2 atomi C

nH = %H*M/AH*100 = 6,96*26/100 = 2 atomi H

F.M [A] - etina (acetilena) C2H2

1 mol (CnH2n).......................14n+160g
0,1 moli...................................20,2g

20,2=1,4n+16 => 4,2=1,4n => n=3

F.M => C3H6 -> propena [B]
3CH₃-CH=CH₂ + 5K₂Cr₂O₇ + 20H₂SO4 -> 3HOOC-CH₃ + 3CO₂ + 23H₂O + 5K₂SO₄ + 5Cr₂(SO₄)₃


C⁻¹ -4e⁻ ⇒ C⁺³ se oxideaza (agent reducator)     ||6||3
C⁻² -6e⁻ ⇒ C⁺⁴ se oxideaza (agent reducator)     ||6||3
Cr₂⁺⁶ +6e⁻ ⇒ Cr₂⁺³ se reduce (agent oxidant)     ||10||5

n=m/M=8,4/42 = 0,2 moli C3H6

3 moli C3H6....................5 moli K2Cr2O7
0,2 moli C3H6.................x=0,333 moli K2Cr2O7

Cm=n/Vs => Vs=0,333/0,1= 3,333 L sol K2Cr2O7

Nu stiu sigur daca am facut bine reactia de oxidare..