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Question

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


Solution

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

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