Question

1. One mole of an ideal gas occupies 22.4 L at STP 

(a) Calculate the mass of 11.2 L of oxygen gas at STP. 

(b) Calculate the number of atoms present in the above sample. 

2. 21 g of nitrogen gas is mixed with 5 g of hydrogen gas to yield ammonia according to the equation. 

N₂ + 3H₂ → 2NH₃ 

Calculate the maximum amount of ammonia that can be formed.

Answer 1

1. (a) Mass of 22.4 L oxygen at STP = 32 g 

∴ Mass of 11.2 L oxygen at STP = 16 g 

(b) No. of atoms present in 16 g of O₂ \(\frac{6.02\times10^{23}}{2}\) ×2 = 6.02 × 10²³ atoms 

2. N₂ + 3H₂ → 2NH₃ 

1 mole N₂ + 3 mole H₂ → 2moles NH₃ 

1 mole N₂ requires 3 mol H₂ 

i.e., 28g N₂ requires 6 g H₂ 

Hence, 21 g N₂ requires \(\frac{6\times21}{28}\) = 4.5 g H₂ 

21g N₂ reacts completely and 0.5g H₂ remains unreacted. 

Hence, N₂ is the limiting reagent.

28g N₂ gives 2 x 17g NH₃

∴ 21 g N₂ gives \(\frac{2\times17\times21}{28}\) = 25.5 g NH₃