Question

68) 2
73) 3,5
78) 4,5
79)1

Answer 1

V68. 2)
fie raportul volumetric=x:y
22,4ml...................67,2ml
C3H8 + 5O2 -> 3CO2 + 4H2O
x.........................3x
22,4ml....................89,6ml
C4H10 + 6,5O2 -> 4CO2 + 5H2O
y..............................4y
x+y=80 si 3x+4y=3(x+y)+y=300 => 240+y=300 => y=60ml butan
x=80-60=20ml propan
%propan=20*100/80=25% si %butan=60*100/80=75%
V73.3)
Cm=n/Vs
n=m/M=1,17g/26g/mol=0,045moli acetilena
Cm=n/Vs=0,045moli/1l=0,045M
5)
crestere masei=masa de propena, caci doar ea se introduce in vas
amestec echimolecular => nr egali de moli => x:x
n=V/Vm=89,6/22,4=4mmoli
2x=4 => x=2mmoli
n=m/M => m=n*M=2*56=112mg=0,112g
cresterea=0,112g sau 112mg
V78.4)
alchena: CnH2n
Cm=n/Vs => n=Cm*Vs=0,05*0,4=0,02moli
14n........1mol
CnH2n + Br2 -> CnH2nBr2
1,12g.....0,02moli
14n=1,12/0,02=56 => n=4
f.m: C4H8
CH3-CH=CH-CH3 + HBr -> CH3-CHBr-CH2-CH3
5)
20% din 5,6=20*5,6/100=1,12m³ metan necesar
22,4m³............1kmol
CH4 + 3Cl2 -> CHCl3 + 3HCl
1,12m³...........x=0,05kmoli
n=0,05kmoli=50moli triclorometan (cloroform)
V79.1)
raport de 1:2:5 => x:2x:5x
1mol...1mol...1mol
C2H2 + H2 -> C2H4
2x.........2x......2x
1mol.....1mol
C2H4 -> C2H4
x............x
1mol..1mol
H2 -> H2
3x.....3x
amestecul initial: 8x iar cel final: 6x
cum rapoartele molare=rapoartele volumetrice, intotdeauna:
=> 8x=179,2=> x=22,4l
Vfinal=6*22,4=134,4l
%etina=2*22,4*100/134,4=33,33%
%etena=22,4*100/134,4=16,66%
%hidrogen=3*22,4/134,4=50%