Question

What volume of nitrogen (N₂) and oxygen (O₂) measured at S.T.P. will be required to prepare 4.6 g of nitrogen dioxide?

Answer 1

Reaction: \[\ce{N2 + 2O2 -> 2NO2}\]

Molecular mass of NO₂ = 14 + 2(16)

= 14 + 32

= 46 g

Moles of NO₂ = `4.6/46`

= 0.1 mole

 From the equation:

Volume of N₂ = `0.1/2 xx 22.4`

= 1.12 L

Volume of O₂ = 0.1 × 22.4

= 2.24 L