Question

20.0 kg of N2(g) and 3.0 kg of H2(g) are mixed to produce NH3(g). The amount of NH3(g) formed is

Answer 1

Answer:

Explanation:

The balanced reaction of ammonia is:

N₂ (g) + 3H₂(g) ⇆ 2NH₃ (g)

In this reaction, it is clear that one mole of N₂ will react with 3 moles of H₂ to form two moles of NH₃

Now, Molar mass of N₂ = 28 g/mole

molar mass of H₂ = 2 g/mole

molar mass of NH₃ = 17 g/mole

No. of moles of N₂ = $\frac{20x10^3}{28}$ ≈ 714 moles (approx)

No. of moles of H₂ = $\frac{3x10^3}{2}$ = 1500 moles of H₂

714 moles of N₂ will react with 2142 moles of H₂, but here, since 1500 moles are available, the limiting reagent will be H₂

So, let x be no of N₂ moles reacting with 1500 moles of H₂ be 'X'

XN₂  + 3XH₂ ⇆ 2XNH₃

in hydrogen, 1500 moles = 3X, then X= 500

now, amount of N₂ molecules reacting will be 500 moles

and amount of ammonia produced will be 1000 moles

Amount of ammonia produced in Kg wil be: 17g x 1000 = 17000g= 17Kg

please mark me brainliest