Question

In a combination reaction, if 1.34794 moles of magnesium is heated with 0.2638 moles of nitrogen, to form magnesium nitride, how many grams of magnesium nitride will be produced?

Answer 1

The balanced equation of combustion of magnesium in nitrogen is

3 Mg + N₂ Mg₃N₂

From the above equation it is clear that 1 mole nitrogen will react completely with 3 moles of magnesium.

We have here 0.2638 moles of nitrogen.

If 1 mole nitrogen will react completely with 3 moles of magnesium, then 0.2638 moles of nitrogen will completely react with (3 x 0.2638) = 0.7914 moles magnesium.

We have therefore in excess of (1.34794 - 0.7914) = 0.55654 moles magnesium

Therefore the limiting reagent here is N₂

Now, 1 mole of N₂ will produce 1 mole of Mg₃N₂

Therefore, 0.2638 moles of N₂ will produce = 0.2638 moles of Mg₃N₂

Molar mass of Mg₃N₂ = 3 x 24.3 + 2 x 14 = 100.9 g/mol

Therefore the amount of Mg₃N₂ will be produced = 0.2638 x 100.9 = 26.62 g