Question

A 9.03 g sample of magnesium reacts completely with 3.48 g of nitrogen to form a compound of
magnesium and nitrogen. What will be the mass of the compound that is formed?

Answer 1

Answer:

12.5 g

Explanation:

12.5 g of the compound would be formed.

First, let us look at the balanced equation of reaction.

$3 Mg(s) + N_2(g) --> Mg_3N_2(s)$

3 moles of Mg is required to react with 1 mole of N2 to produce 1 mole of product.

Recall that: mole = mass/molar mass

9.03 g of Mg = 9.03/24.3 = 0.3716 mole

3.48 g of N2 = 3.48/28 = 0.1243 mole

Mole ratio of Mg/N2 = 3:1

Hence, there is no limiting reactant.

3 moles of Mg is required for 1 mole of product.

0.3716 mole of Mg will therefore require:

       0.3716 x 1/3 = 0.1239 moles of product.

Molar mass of product $Mg_3N_2$ = 100.9 g/mol

Mass of 0.1239 mole $Mg_3N_2$ = mole x molar mass

                      = 0.1239 x 100.9 = 12.5 g