How many litres of oxygen measured at stp required for complete combustion of 39 gram of liquid cs6
Answers
Your answer is here :
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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.
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