# 4.43 Magnesium nitride forms in a side reaction when magnesium metal burns in air. This reaction may also be carried out in pure nitrogen. 3 Mg ( s ) + N 2 ( g ) → Mg 3 N 2 ( s ) If 18.4 g of Mg 3 N 2 forms from the reaction of 20.0 g of magnesium with excess nitrogen, what is the percentage yield?

### Chemistry for Engineering Students

3rd Edition
Lawrence S. Brown + 1 other
Publisher: Cengage Learning
ISBN: 9781285199023

### Chemistry for Engineering Students

3rd Edition
Lawrence S. Brown + 1 other
Publisher: Cengage Learning
ISBN: 9781285199023
Chapter 4, Problem 4.45PAE
Textbook Problem
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## 4.43 Magnesium nitride forms in a side reaction when magnesium metal burns in air. This reaction may also be carried out in pure nitrogen.   3 Mg ( s )   +  N 2 ( g )   →  Mg 3 N 2 ( s ) If 18.4 g of Mg3N2forms from the reaction of 20.0 g of magnesium with excess nitrogen, what is the percentage yield?

Interpretation Introduction

Interpretation:

To calculate the percentage yield.

Concept introduction:

• The maximum yield is obtained by reacting all the limiting reactant.
• Percent yield relates actual versus theoretical yields.

Given:

mass=18.4gMg2N3MolecularWeightofMg3N2=100.93g/molmass=20.0gMgAtomic Weight=24.0u

### Explanation of Solution

Use the given amounts to identify the limiting reactant. Then calculate the total amount of possible iron yield.

Step 1: Balance Reaction

Reaction is balanced

Step 2: Change all mass units to mol units

molMg3N2=massMolecular Weight=18.4100.9284=0.184molMg=massMolecular Weight=20.024.0=0.833

Step 3: Relate stoichiometry

Ratio=1:3molMg3N2=13×(0

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