Question

10g of hydrogen and 64g of oxygen were filled in a steel vessel and exploded.vol. of gaseous product obtained

Answer 1

The balanced equation for this reaction is as follows: 

                   2H₂ + O₂ -------> 2H₂O

Find the moles of the reacting gases.

10g of hydrogen

moles = mass/molar mass
          = 10g / 2
          =  5 moles

64g of oxygen

mass/molar mass

moles = 64/32

 = 2 moles

The mole reaction for the equation above is H₂ : O₂  is 2 :1

Therefore 5 moles of H₂ will require 2.5 moles of oxygen - this makes oxgen the limiting reagent.

We therefore use the moles of oxygen to find those of H₂O:

mole ratio  O₂ : H₂O   is 1 : 2

That means 2 moles of oxygen will produce 4 moles of H₂O

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Calculate volume of 4 moles of H₂O gas.

Ideal gas law states that 1 mole of an ideal gas occupies exactly 22.4 liters by volume.

If 1 mole = 22.4 liters
Then 4 moles = 22.4 × 4/1
                       = 89.6 liters

The volume of H₂O produced is 89.6 liters.