Answers

2014-08-30T23:22:04+05:30
Let no. of mole of C2H4 is n
pressure = 380/760 atm = 1/2 atm
T = 273+273 = 546K
V = 11 L
R = 0.0821
apply 
PV = nRT
11/2= n*0.0821*546
n = 0.123 mole
for one mole of combustion of C2H4 3mole of O2 is required
so for 
0.123 mole = 3*0.123 mole  O2 required
                = 0.369mole
volume of one mole gas at stp = 22.4L
volume of 0.369 mole O2 at stp = 22.4*0.369 = 8.27L
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