Question

Unde gresesc?La sfarsit da alte procente.

Answer 1

a)
Raport de 1:2:2 => x moli metan, 2x moli etena si 2x moli propena
1mol......1mol....1mol
C2H6 -> C2H4 + H2
x moli....x moli...x moli
1mol.....1mol.....1mol
C3H8 -> C3H6 + H2
2x moli..2x moli..2x moli
1mol......1mol.....1mol
C3H8 -> C2H4 + CH4
x moli....x moli...x moli
intial: 4x moli amestec etan si propan
final: 8x moli amestec metan, etena, propena si H2
%etan=x*100/4x=25%
%propan=3x*100/4x=75%
Mmedie=(16x+28*2x+42*2x+2*3x)/(8x)=(16x+56x+84x+6x)/8x=162x/8x=20,25g/mol
daer=Mamestec/Maer=20,25/28,9=0,7
b)
raport etan: propan=x:3x=1:3
n=m/M
pVM=mRT => pM=(m/V)RT=dRT
M=(30x+44*3x)/4x=162/4x=40,5g/mol
d=pM/RT=40,5/0,082*300=1,64g/l
c)
din 3x moli propan, 2x se dehidrogeneaza
randament=2x*100/3x=66,66%
d)
p1V1=n1RT
p2V2=n2RT
le impartim => p1V1/p2V2=n1/n2
insa consideram raportul volumelor constant => p1/p2=n1/n2 => p1/p2=4x/8x => p1/p2=1/2 => p2=2p1 => presiunea creste de 2 ori
e)
n=V/Vm => V=n*Vm
Vamestecinitial=n*Vm=22,4*4x=89,6x m³
VH2=22,4*3x=67,2x m³
89,6x m³ amestec...........67,2x m³ H2
8 m³ amestec.........................6 m³ H2
95/100=Vpractic/6 => Vpractic=6*95/100=5,7m³ H2 separati
98/100=5,7/Vimpur => Vimpur=5,7*100/98=5,81m³ H2 impuri