Answer to Question #60338 in General Chemistry for Hafsa

Question

The following is a Limiting Reactant problem:
Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g)  Mg3N2(s)
How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)? Show all calculations leading to an answer.

Answer 1

As one mole of N2 requires 3 moles of magnesium to react completely, 2 moles of N2 require 6 moles of Mg. According to the problem statement, 8 moles of magnesium are given, it is greater that the amount required, therefore, Mg is in abundance and nitrogen is the limiting reactant.
One mole of nitrogen leads to one mole of the product, and, proportionally, two moles of nitrogen (n(N2) = 2 mol) lead to two moles of the product (n(Mg3N2)= n(N2) = 2 mol). The molar mass of magnesium nitride is M(Mg3N2)=100.93 g/mol. Therefore, the mass of the final product is
m=n(Mg3N2)M(Mg3N2)=(2 mol)×(100.93 g/mol) = 201.86 g.
Answer: 201.86 g.