Question

Ammonia can be made by reaction of water with magnesium nitride as shown by the following unbalanced equation: Mg3N2(s) + H2O(l)  Mg(OH)2(s) + NH3(g) If this process is 71% efficient, what mass of ammonia can be prepared from 24.5 kg magnesium nitride?

Answer 1

The mass of ammonia prepared from 24.5 kg magnesium nitride, according to the reaction Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g), knowing that the process is 71% efficient is 5.87 kg.

The balanced reaction of production of ammonia is:

Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g) (1)

First, let's find the number of moles of magnesium nitride

`n_{Mg_(3)N_(2)} = \frac{m_{Mg_(3)N_(2)}}{M_{Mg_(3)N_(2)}}` (2)

Where:

`m_{Mg_(3)N_(2)}`: is the mass of Mg₃N₂ = 24.5 kg

`M_{Mg_(3)N_(2)}`: is the molar mass of Mg₃N₂ = 100.9494 g/mol

The number of moles is (eq 2):

`n_{Mg_(3)N_(2)} = \frac{m_{Mg_(3)N_(2)}}{M_{Mg_(3)N_(2)}} = (24500 g)/(100.9494 g/mol) = 242.70 \:moles`

We can calculate the mass of ammonia prepared, knowing that 1 mol of Mg₃N₂ reacts with 6 moles of H₂O to produce 3 moles of Mg(OH)₂ and 2 moles of NH₃ (reaction 1).

`n_{NH_(3)} = (2\: moles\: NH_(3))/(1\: mol\: Mg_(3)N_(2))*n_{Mg_(3)N_(2) = (2\: moles\: NH_(3))/(1\: mol\: Mg_(3)N_(2))*242.70 \:moles \:Mg_(3)N_(2) = 485.4 \:moles`

Then, the mass of NH₃ is:

`m_{NH_(3)} = n_{NH_(3)}*M_{NH_(3)} = 485.4 \:moles*17.031 g/mol = 8266.8 g = 8.27 kg`

Since the process is 71% efficient, the mass that can be prepared is:

`m = 8.27 kg*0.71 = 5.87 kg`

Therefore, the mass of ammonia that can be prepared is 5.87 kg.

I hope it helps you!