Chapter 12, Problem 3STP

Interpretation Introduction

Interpretation: The correct option in accordance with the reaction of 6.0 mol of magnesium and 4.0 mol of nitrogen is to be stated.

Concept Introduction: According to the law of conservation of mass, the mass can neither be created nor be destroyed. Every reaction must be balanced. With the help of the balanced equation, the limiting reagent can be found and in accordance with it, the amount of product can be calculated.

Answer to Problem 3STP

The correct option is (B).

Explanation of Solution

Given information:

• 4.0 mol of nitrogen
• 6.0 mol of magnesium

The balanced reaction is given as follows:

  $\text{3Mg}\left(s\right)+{\text{N}}_2\left(g\right)\to {\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\left(s\right)$

According to the reaction,

  $\begin{array}{c}\text{3}\text{mol}\text{of}\text{Mg}\to 1\text{mol}\text{of}{\text{N}}_2\\ \text{1}\text{mol}\text{of}\text{Mg}\to \frac{1}{3}\text{mol}\text{of}{\text{N}}_2\\ \text{6}\text{mol}\text{of}\text{Mg}\to \frac{1}{3}\times 6\text{mol}\text{of}{\text{N}}_2\\ \to 2\text{mol}\text{of}{\text{N}}_2\end{array}$

Thus, the nitrogen present in excess which is 4.0 − 2.0 = 2.0 mol of nitrogen is extra. The limiting reagent is Mg so the mol of the product is calculated as follows:

  $\begin{array}{c}\text{3}\text{mol}\text{of}\text{Mg}\to 1\text{mol}\text{of}{\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\\ \text{1}\text{mol}\text{of}\text{Mg}\to \frac{1}{3}\text{mol}\text{of}{\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\\ \text{6}\text{mol}\text{of}\text{Mg}\to \frac{1}{3}\times 6\text{mol}\text{of}{\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\\ \to 2\text{mol}\text{of}{\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\end{array}$

Thus, 2 mol of ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$ is formed. Hence, the correct option is (B).