Answer to Question #100052 in General Chemistry for Malathy

Question

Magnesium reacts directly with nitrogen gas to form magnesium nitride. in an experiment conducted, it was found that 0.36g of magnesium produced 0.50g of magnesium nitrite.

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 nitrite

Answer 1

As 0.36 g of magnesium produced 0.50 g of magnesium nitride, the number of moles are:

n(magnesium nitride) = 0.50 g / 101 g/mol = 0.005 mol

n(magnesium) = 0.36 g / 24 g/mol = 0.015 mol

a) As 0.005 mol of magnesium nitride was produced, it can be concluded, that 0.005 moles of nitrogen gas and 0.01 moles of nitrogen combined with 0.36 g of magnesium.

b) The numbers of nitrogen atoms and magnesium atoms in magnesium nitride are:

N(magnesium) = n(magnesium) / n(magnesium nitride) = 0.015 mol / 0.005 mol = 3

N(nitrogen) = n(nitrogen) / n(magnesium nitride) = 0.01 mol / 0.005 mol = 2

As a result, the ratio between magnesium atom to nitrogen atoms in magnesium nitride is 3/2.

c) As the ratio between atoms is 3/2, the empirical formula of magnesium nitride is Mg₃N₂.

Answer: a) 0.005 mol of nitrogen gas; b) the ratio is 3/2; c) empirical formula is Mg₃N₂.