Question

Given that 45.0 mL of nitrogen gas reacts with 65.0 mL of oxygen gas according to the reaction:

N2 (g) + 2 O2 (g) = 2 NO2 (g)

a) What is the limiting reactant?

b) Assuming constant conditions, what is the volume of NO2 produced?

Answer 1

a)

N₂ (g) + 2 O₂ (g) = 2 NO₂ (g)

From the above reaction volume ratio of N₂ and O₂ is 1:2

Volume of O₂ required to react with 45 mL N₂ =

= 90 mL O₂

But we have only 60 mL of O₂. So O₂ voume is less

Therefore limiting reactant is O₂

b)

Volume of NO₂ produced =

= 60 mL NO₂

Volume of NO₂ produced = 60 mL