Find out the amount of oxygen required to burn 39g benzene completely.?

The equation of benzene reacting with oxygen is
2C6H6 +15O2= 12CO2 +6H2O
The molar weight of 2C6H6 is 156gm
The molar weight of 15O2 is 480gm
At that point for ignition of 39gm of C6H6, O2 required =480 *39/156
=120gm of O2
In this manner the volume of 120gm of O2 at STP = 22.4 *(120/32 )
Since (molar volume of O2 =22.4L)
=84Liters