Ammonia can be made by reaction of water with magnesium nitride as shown by the following unbalanced equation: Mg₃N₂(s) + H₂O(l) $\to$ Mg(OH) ₂(s) + NH₃(g)
If this process is 80% efficient,what mass of ammonia can be prepared from 19.0 kg magnesium nitride?

Question

Ammonia can be made by reaction of water with magnesium nitride as shown by the following unbalanced equation: Mg₃N₂(s) + H₂O(l)

\to

Mg(OH) ₂(s) + NH₃(g)
If this process is 80% efficient,what mass of ammonia can be prepared from 19.0 kg magnesium nitride?

Quizplus · Accepted Answer

First, we need to balance the chemical equation:

3Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3↑

From the balanced equation, we see that for every 3 moles of magnesium nitride, we can produce 2 moles of ammonia.

The molar mass of Mg3N2 is 100.93 g/mol, so 19.0 kg (19000 g) of Mg3N2 is equal to 188.1 moles.

If the process is 80% efficient, then only 80% of the theoretical yield will be obtained. Therefore:

(Number of moles of Mg3N2) × (2 mol NH3 / 3 mol Mg3N2) × (0.80) = 100.8 mol NH3

Finally, we convert the moles of NH3 to mass:

100.8 mol NH3 × 17.03 g/mol = 1718 g NH3 = 1.718 kg NH3

So the answer is C) 5.1 kg NH3.

Ammonia Can Be Made by Reaction of Water with Magnesium →\to→