Question

benzene burns in oxygen according to equation to C6 H6 + 15 O2 gives 12C
O2+6 H2Ohow many litres of oxygen are required at STP for complete combustion of 39 gram of liquid Benzene ​

Answer 1

Answer:

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.