A mixture of C2H4 and C3H8 was kept in a 0.820L vessel at 1 atm and 300 K. The weight of the gas mixture in the mixture 0.613g. Calculate the ratio of moles of C3H8 and C2H4.
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Explanation:
For O
2
, PV=nRT
or, 1×40=n×0.0821×400
or, n=1.218
C
2
H
6
(g)+
2
7
O
2
(g)→2CO
2
(g)+3H
2
O(l)
C
2
H
4
(g)+3O
2
(g)→2CO
2
(g)+3H
2
O(l)
Let the moles of ethane be a.
Volume of O
2
required by ethane =7/2a
Volume of ethane =(1.218−a)
Volume of O
2
required by ethene=3(1.218−a)
Given,
2
7
a+3(1.218−a)=
32
130
(moles of oxygen)
or
2
7
a+3.654−3a=4.0625
or a=0.817(moles of ethane)
∴1.218−a=1.218−0.817 =0.401(moles of ethene)
Mole fraction of ethane =
1.218
0.817
=0.6707
Mole fraction of ethene=1−0.6707=0.3293
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