Question

448 L of Ammonia at STP on decomposition gives Xg of Nitrogen. How many molecules of Oxygen are required to produce NO₂ from Xg of Nitrogen?

Answer 1

• According to Avogadro's Law, equal volumes of any gas have an equal number of molecules at the same temperature and pressure. one mole of gas contains 6.023×10²³ molecules.
• Chemical equation for the decomposition of Ammonia

2NH₃ (g)⇢ N₂ (g)+ 3H₂(g)

• From the equation

2 × 22.4L = 44.8 L of Ammonia produces 22.4L of Nitrogen N₂

Therefore 448 L of Ammonia produces 224L of Nitrogen(N₂).

• Molecular weight of 22.4L of Nitrogen =2×14=28g

Therefore 224L of Nitrogen has a weight= $\frac{28\times 224}{22.4}$=280g

So X=280g

• Chemical equation for the formation of Nitrogen dioxide

N₂ (g)+ 2O₂(g)⇢2NO₂(g)

28g of N₂ ⇢ 64g of O₂

Therefore 280g of N₂⇢ 640 g of O₂

32 g of Oxygen contains molecules= 6.023×10²³

Therefore 640g of O₂ contains= $\frac{640\times 6.023\times {10}^{23}}{32}$= 12×10²⁴

Hence, the number of molecules of Oxygen required is 12×10²⁴.