Question

volume of o2 and o3 obtained by 8 kg oxygen atoms at STP

Answer 1

Mass of 1 oxygen atom = 16 g.
Mass of O₂ = 32 g.
Mass of O₃ = 48 g.

Now,
16 g of Oxygen atom combines to form 32 g of O₂.
∴ 1 g of O atom combines to forms 2 g of O₂.
∴ 8000 g of O atom combines to form 16000 g of O₂.

Similarly, Mass of O₃ (Ozone) formed = 3 ×  8000 g. = 24000 g.

Now,
1 mole of Oxygen at S.T.P. occupies the Volume of 22.4 liter.
∴ 32 g of Oxygen at S.T.P. occupies the volume of 22.4 liter.
∴ 1 g of Oxygen at S.T.P. occupies the volume of 22.4/32 liter.
∴ 16000 g of Oxygen at S.T.P. occupies the volume of 11200 liter.

Hence, the volume of oxygen O₂ at S.T.P. is 11200 liter.

1 mole of Ozone at S.T.P. occupies the volume of 22.4 liter. 
∴ 48 g of Ozone at S.T.P. occupies the volume of 22.4 liter. 
∴ 1 g of Ozone at S.T.P. occupies the volume of 22.4/48 liter. 
∴ 24000 g of Ozone at S.T.P. occupies the volume of 22.4/48 × 24000 liter. 
  = 11200 liter.

Hence, the volume of ozone formed is 11200 liter.


Hope it helps.