6 moles of N2(g) & 9 moles of H2(g) are allowed to react in a closed rigid vessel at TK temperature. Calculate volume in litre of NH3(g)
formed at STP, if % yield of reaction is 50%
Answers
Answer: Bro the answer is :
6 moles of N₂ and 9 moles of H₂ are allowed to react in a closed vessel at a certain temperature.
We have to find the volume in litre of NH₃ formed at STP, if % yield of reaction is 50 %.
Reaction between nitrogen and hydrogen :
it is an association reaction in which nitrogen reacts with hydrogen and produces Ammonia. the balanced chemical reaction of it is given as,
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g),
here you see, one mole of nitrogen reacts with three moles of hydrogen and produces two moles of Ammonia.
given,
no of moles of nitrogen = 6
no of moles of hydrogen = 9
∵ 1 mol of N₂ reacts with 3 mol of H₂.
∴ 6 mol of N₂ will react with 18 mol of H₂. but here only 9 mol of H₂ is available. hence, H₂ is limiting reagent.
so the reaction will proceed according to H₂.
you see, 3 mol of H₂ produces 2 mol of ammonia.
∴ 9 mol of H₂ will produce 6 mol of ammonia.
theoretical value of no of moles of NH₃ = 6
% yield of reaction is 50%.
so the no of moles NH₃ = 50% of 6 = 3 mol
we know,
at STP, 1 mol of any ideal gas occupies 22.4 L volume.
so, the volume of NH₃ at STP = 3 × 22.4 = 67.2 L