[1] calculate the volume of NH3 gas produced at STP when 140 g of N2 reacts with 30 g of H2?
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Moles of N₂ = 140 / 28 = 5 mol
Moles of H₂ = 30 / 2 = 15 mol
Balanced equation of formation of NH₃
N₂ + 3H₂ → 2NH₃
From above balanced equation we can say that for formation of two mole of NH₃ 1 mol of N₂ and 3 mol of H₂ is required.
We have 5 and 15 mol of N₂ and H₂ respectively.
There is no limiting reagent.
Now,
3 mol of H₂ gives 2 moles of NH₃
15 mol of H₂ will give 10 moles of NH₃
Moles of NH₃ produced is 10
At STP
One mole of any gas occupies 22.4 Litre
10 mole of NH₃ at STP will occupy 224 Litre
Volume of NH₃ gas formed is 224 Litres
Moles of H₂ = 30 / 2 = 15 mol
Balanced equation of formation of NH₃
N₂ + 3H₂ → 2NH₃
From above balanced equation we can say that for formation of two mole of NH₃ 1 mol of N₂ and 3 mol of H₂ is required.
We have 5 and 15 mol of N₂ and H₂ respectively.
There is no limiting reagent.
Now,
3 mol of H₂ gives 2 moles of NH₃
15 mol of H₂ will give 10 moles of NH₃
Moles of NH₃ produced is 10
At STP
One mole of any gas occupies 22.4 Litre
10 mole of NH₃ at STP will occupy 224 Litre
Volume of NH₃ gas formed is 224 Litres
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