Question

In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to produce ammonia. The conditions included medium temperature (~500oC), very high pressure (~351kPa), and an iron catalyst. The reaction is represented by the equation:

N2(g) + 3H2(g) ↔ 2NH3(g)


Assuming a similar process takes place at STP, what volume of ammonia gas will be produced if 2.4L nitrogen are reacted with an excess of hydrogen gas?

A)
1.2L NH3


B)
2.4L NH3


C)
4.8L NH3


D)
9.3g NH3

Answer 1

C) 4.8 L NH₃.

Step-by-step explanation:

• The balanced equation for the mentioned reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g),

It is clear that 1.0 moles of N₂ react with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

• At STP, 2,4 L of N₂ reacts with an excess of hydrogen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

• using cross multiplication:

1.0 mol of N₂ represents → 22.4 L.

??? mol of N₂ represents → 2.4 L.

∴ 2.4 L of N₂ represents = (1.0 mol)(2.4 L)/(22.4 L) = 0.1071 mol.

• To find the no. of moles of NH₃ produced:

• Using cross multiplication:

1.0 mol of N₂ produce → 2.0 mol of NH₃, from stichiometry.

0.1071 mol of N₂ produce → ??? mol of NH₃.

∴ The no. of moles of NH₃ = (2.0 mol)(0.1071 mol)/(1.0 mol) = 0.2142 mol.

• Again, using cross multiplication:

1.0 mol of NH₃ represents → 22.4 L, at STP.

0.2142 mol of NH₃ represents → ??? L.

∴ The no. of liters of NH₃ will be produced = (0.2142 mol)(22.4 L)/(1.0 mol) = 4.789 L ≅ 4.8 L.

So, the right choice is: C) 4.8 L NH₃.