biochemical calculation please help me

Question

Find the molar ratio at which N2O and N2O3 were mixed, if we know that 11 dm3 of this mixture reacts with 27 dm3 of hydrogen, and that nitrogen and water are the only products of this reaction. (all volumes are measured at the same temperature and pressure conditions).

Answer 1

If this reaction actually takes place, we could write the following balanced equation:

N₂O + N₂O₃ + 4H₂ ==> 2N₂ + 4H₂O

Since the temperature and pressure are constant, moles and volumes of substances can be used interchangeable. Thus, in this equation, we can say that 1 mol or 1 dm³ of N₂O reacts with 1 mole or 1 dm³ of N₂O₃, etc.

In the balanced equation, the mole ratio (or volume ratio) of N₂O to N₂O₃ is 1:1, but since the question asks for the mole ratio, then one of them must have been present in limiting supply.

We can use the mole ratio (or volume ratio) of H₂ (assuming it was present in excess) to help solve this.

1 mol N₂O + 1 mol N₂O₃ + 4 mol H₂ is the balanced reaction

2 mol mixture / x mol mixture = 4 mol H₂ / 27 mol H₂

x = 13.5 mols of mixture, but problem states it took only 11 mols of mixture

11 / 13.5 = 0.81 moles of either N₂O or N₂O₃

Thus, mole ratio N₂O and N₂O₃ = 0.81:1 or 1.23:1