22,4
C2H4 + H2 -->C2H6
x moli 22,4*x l
C3H6 + H2-->C3H8
y moli 22,4*y l
22,4x+22,4y=44,8 (Volum total de H2) =>x+y=2
masa de amestec=x*M(etena)+y*M(propena)=x*28+y*42=58,4
=>14x+21y=29,2
14x+21*(2-x)=29,2
14x+42-21x=29,2 => 12,8=7x => x=1,828 moli etena<=>1,828*28=51,2 grame de etena
y=2-1,828=0,172 moli propena<=>7,244 g sau y se poate calcula prin 58,4-51,2.
