Question

Liquid benzene (C6H6) burns inoxygen according to 2C6H6(l) +15O2 --->12CO2 + 6H2O how many litre of O2 at STP are needed to complete the combustion of 39g of liquid benzene?

Answer 1

To calculate this question:

Step 1: we need to find out 39g of liquid benzene is equivalent to how many moles of the liquid.

Moles = mass/molar mass
mass = 39g
molar mass= 78

Therefore moles of liquid benzene = 39/78 = 0.5 moles

Step 2: Use the equation plus the moles of the liquid benzene to find the moles of O₂ used.

2C₆H₆ +15O₂ --->12CO₂ + 6H₂O

We see that for every 2 moles of liquid benzene we require 15 moles of O₂ for the reaction. Thus;

If 2 moles benzene needs 15 moles of oxygen,
Then 0.5 moles will need   15 ₓ 0.5/2 = 3.75 moles of oxygen.

Step 3: Calculate the volume of oxygen required from its moles calculated
 
The ideal gas law states that 1 mole of an ideal gas will occupy 22.4 liters at STP(standard temperature and pressure)

Therefore if 1 mole of oxygen occupies 22.4 liters,
Then 3.75 moles of oxygen will occupy;
                       22.4 liters ₓ 3.75 moles/1 mole = 84 liters

Therefore 84 liters of oxygen is what is required to completely burn 39g of liquid benzene. 

Answer 2

2 C6H6 + 15 O2 --------------------> 12CO2 + 6 H2O

156gm -----480gm

so for combustion of 39 gm of C6H6 O2 required = 480x39/156

= 120 gm O2

so volume of 120 gm of O2 at STP = 22.4 x120/32 [ molar volume of O2= 22.4 L ]

= 84 liters answer