Question

Magnesium reacts directly with nitrogen gas to form Magnesium Nitride. In an experiment conducted, it was found that 0.36g of magnesium produced 0.50 magnesium nitride.
(a)How many moles of nitrogen combined with 0.36g of Magnesium.
(b)What is the ratio of nitrogen atoms to magnesium atoms in magnesium nitride.
(c)What is the empirical formula of magnesium nitride?

Answer 1

3Mg + N₂ → Mg₃N₂

(a) $n(Mg) = \frac{0.36}{24} = 0.015 \;mol$

$n(N_2) = \frac{1}{3}n(Mg) = \frac{0.015}{3} = 0.005 \;mol$

(b) N atoms : Mg atoms = 2 : 3

(c) Mg₃N₂