magnesium nitride and water react to produce a precipitate of magnesium hydroxide and ammonia gas. Mg3N2+ 6H2O-3Mg(OH)2+ 2NH3. How much magnesium nitride is needed to produce 87g of the precipitate in excess water?

Respuesta :

The balanced reaction

Mg₃N₂ + 6 H₂O   →   3 Mg(OH)₂ + 2 NH₃

informs you that you need 1 mol magnesium nitride (Mg₃N₂) to produce 3 mol magnesium hydroxide (Mg(OH)₂).

Convert 87 g of Mg(OH)₂ to moles (molar mass ≈ 58.319 g/mol) :

(87 g) (1/58.319 mol/g) ≈ 1.4918 mol

We need 1/3 of this amount of Mg₃N₂, or about 0.49727 mol. Convert this to grams (molar mass ≈ 100.93 g/mol) :

(0.49727 mol) (100.93 g/mol) ≈ 50.19 g ≈ 50. g