Question

What amount of magnesium nitride is required to react completely with 0.25 mol of water? What amounts of the products are expected?

Answer 1

Mg₃N₂ + 6H₂O --> 3Mg(OH)₂ + 2NH₃

According to the equation, the amounts in moles are:

n (Mg₃N₂) = 0.25 * 1/6 = 0.042 mol

n (Mg(OH)₂) = 0.25 * 3/6 = 0.13 mol

n (NH₃) = 0.25 * 2/6 = 0.083 mol

The masses are:

m (Mg₃N₂) = 0.25 * 1/6 * 100.95 = 4.2 g

m (Mg(OH)₂) = 0.25 * 3/6 * 58.32 = 7.3 g

m (NH₃) = 0.25 * 2/6 * 17.03 = 1.4 g