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

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Answer 2

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):

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

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).

$n_{NH_{3}} = \frac{2\: moles\: NH_{3}}{1\: mol\: Mg_{3}N_{2}}*n_{Mg_{3}N_{2} = \frac{2\: moles\: NH_{3}}{1\: mol\: Mg_{3}N_{2}}*242.70 \:moles \:Mg_{3}N_{2} = 485.4 \:moles$

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!