Mole ConceptHard
Question
A mixture of ethylene and excess of H2 had a pressure of 600 mm Hg. The mixture was passed over nickel catalyst to convert ethylene to ethane. The pressure of the resultant mixture at the similar conditions of temperature and volume dropped to 400 mm Hg. The fraction of C2H4 by volume in the original mixture is
Options
A.1/3 rd of the total volume
B.1/4 th of the total volume
C.2/3 rd of the total volume
D.1/2 of the total volume
Solution
Let there be n mole of (C2H4 + H2)
Mole of C2H4 = x
H2 = (n - x) mole
C2H4 + H2 → C2H6
Mole of C2H4 = x
H2 = (n - x) mole
C2H4 + H2 → C2H6
x n-x
0 n - 2x x mol
After reaction (C2H6 + H2 left)
x + n - x - x = n - x
[Total H2 = (n - x), H2 reacted = x]
H2 left = (n - x - x)
n ∝ 600; n - x ∝ 400
;
0 n - 2x x mol
After reaction (C2H6 + H2 left)
x + n - x - x = n - x
[Total H2 = (n - x), H2 reacted = x]
H2 left = (n - x - x)
n ∝ 600; n - x ∝ 400
; x/n = 1/3 = fraction of C2H4 by volume
Create a free account to view solution
View Solution FreeMore Mole Concept Questions
The density (in g/L) of an equimolar mixture of methane and ethane at 1 atm and 0o C is...Law of multiple proportions is not applicable for the oxide(s) of...Accumulation of which of the following molecules in the muscles occurs as a result of vigorous exercise ?...Law of multiple proportions is not applicable for the oxide(s) of...1.2575 g sample of [Cr(NH3)6]SO4Cl(Mw = 251.5) is dissolved to prepare 250mL solution showing and osmotic pressure of 1....