pornim de la 1 mol de propan. notam a moli dehidrogenat si b moli ramasi netransformati
1mol 1mol 1mol
C₃H₈ -> C₃H₆ + H₂
a moli a moli a moli
1mol 1mol
C₃H₈ -> C₃H₈
b moli b moli
Mamestec=(npropena*Mpropena+nhidrogen*Mhidrogen+npropan*Mpropan)/(namestec)
=> 27,5*(2a+b)=42a+2a+44b => 55a+27,5b=44a+44b => 11a=16,5b => a=1,5b
insa: a+b=1 mol
1,5b+b=1=2,5b => b=0,4moli deci a=0,6moli
1mol..........100%
0,6moli........60%
randamentul=60%
