Question

What volume of oxygen at STP is required to affect combustion of 11litres of ethlene at 273centigrede and 380 mm of Hg pressure

Answer 1

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

Answer 2

Answer:8.27L

Explanation:

Let n, be the no of moles of C2H4

Pressure =380/760=1/2atm

Temperature =263+273=546K

Volume =11L

Now,

Apply PV=nRT;

11/2=n*0.821*548

n=0.123mole

For one mole of combustion of C2H4 3mole of oxygen is required

Therefore,

For 0.123mole =3*0.123mole

=0.369mole of oxygen is required.

Volume of mole gas at STP =22.4L

Volume of 0.369 mole of oxygen at STP is equal to 22.4*0.369=8.27L