Question

A 21.496 g sample of magnesium is burned in air to form magnesium oxide and magnesium nitride. When the products are treated with water, 2.813 g of gaseous ammonia are generated. Calculate the amounts of magnesium nitride and magnesium oxide formed.

Answer 1

The equations involved in this reaction are;

2Mg (s) + O2 (g) -----> 2MgO (s)

3Mg (s) + N2 (g) -----> Mg3N2 (s)

Mg3N2 (s) + 6H2O (l) -----> 3Mg(OH)2 (s) + 2NH3 (g)

The molar mass of NH3 is 17.0 g/mol.

$2.813 g of NH3 * (1 mol / 17.0 g) = 0.165 moles$ of NH3

2 mol NH3 are produced from 1 mol Mg3N2.

$0.165 moles of NH3 x (1 mole of Mg3N2 / 2 moles of NH3) = 0.0827 moles$ of Mg3N2

The molar mass of Mg3N2 is 100.9 g/mol.

$0.0827 moles of Mg3N2 x (100.9 g / 1 mole) = 8.34 g of Mg3N2.$

$0.248 mol Mg x (24.3 g / 1 mol) = 6.03 g of Mg$ (present in the Mg3N2)

$21.296 g - 6.03 g = 15.3 g of Mg$(present in MgO)

1 mole MgO contains 24.3 g of Mg and 16.0 g of O for a total of 40.3 g.

$15.3 g Mg x (40.3 g MgO / 24.3 g Mg) = 25.3 g of MgO$