In a combination reaction, if 1.34794 moles of magnesium is heated with 0.2638 moles of nitrogen,...

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?

Expert Solution

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

queen_honey_blossom answered 2 years ago

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