Răspun
niu.am= V/Vm= 11,2/22,4 mol=0,2mol
amestecul contine alchena A CnH2n SI alchena B-Cn'H2n'
niu,A/niuB=4/1 niuA+niuB=0,5---> niuA=0,4mol niuB=0,1mol
n/n'=2/3
din compozitia produsilor de ardere, deduc ca o alchena are 4atC si o ramificare
deci n= 4 ---> izobutena CH2=C(CH3)-CH3
iar n'=6---> ca sa rezulte acetona la ardere, alchena este
CH3-C(CH3)=C(CH3)-CH3--2,3 dimetil 2 butena
verificare:
REACTIILE DE ARDERE SUNT :
CH2=C(CH3)-CH3+4O----> CO2+H2O+CH3-CO-CH3
0,4mol.....................1,6mol...0,4mol..0,4mol........0,4mol.
CH3-C(CH3)=C(CH3)-CH3+2O----> 2CH3-CO-CH3
0,1mol..........................................0,2mol....0,2mol
total: 0,6mol acetona, 0,4molCO2, 0,4molH2O in raport 3;2:2
total : 1,8molO
c,m= niu/V---> niu= 0,72molKMnO4
2KMno4 + 3H2SO4 ---> K2SO4 + 2MnSO4 + 3H2O+ 5O
2mol........................................................................................5mol
0,72mol..........................................................................x x= 1,8molO
Explicație:
