Answer to Question #68925 in Physics for DEEPENDRA SINGH

Question

a ideal mixture of C2H6 and C2H4 covered 28 leter volume at 0 degree centigrate and 1 atm .mixture react with 128 gram of oxygen gass and make CO2 and H2O.what is the mole fraction of C2H6?

Answer 1

PV = nRT
n=PV/RT=((101 325)(0.028))/((8.31)(273.15))=1.25 mol
Let mole of C_2 H_6 and C_2 H_4 is a and b respectively.
Therefore,
a + b = 1.25→b= 1.25-a
Now,
C_2 H_6+7/2 O_2→2CO_2+3H_2 O
C_2 H_4+3O_2→2CO_2+2H_2 O
Thus, number of moles of O_2 needed for complete reaction of mixture is
n(O_2 )=7a/2+ 3b.
n(O_2 )=m(O_2 )/M(O_2 ) =128/32
7a/2+ 3b=128/32
Thus,
7a/2+ 3(1.25-a)=128/32
a=0.5 mol
Therefore, mole fraction of C_2 H_6 is
0.5/1.25=0.4.