Chapter 13, Problem 41A

(a)

Interpretation Introduction

Interpretation: - The balanced chemical equation when Sodium metal reacts with nitrogen gas to produce sodium nitride is to be written.

Concept Introduction: -

Equation is balanced when number of atoms of each element on both sides is equal.

Answer to Problem 41A

  $6Na+N_2\to 2N{a}_{3}N$

Explanation of Solution

When sodium metal reacts with ${\mathrm{N}}_2$ gas it results in the formation of ${\mathrm{Na}}_{3}N$ .

Balanced chemical equation

  $\begin{array}{l}{\text{6Na + N}}_{\text{2}}\to {\text{ 2Na}}_{\text{3}}\text{N}\\ \text{sodium nitrogen sodium}\\ \text{ gas nitride}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The amount of sodium nitride produced is to be calculated.

Concept introduction:

The amount of sodium nitride produced can be calculated by using formula.

Number of moles = $\frac{\text{mass}}{\text{molar mass}}$

  $\text{mass = no}\text{. of moles }\times \text{ molar mass}$

Answer to Problem 41A

165.9g of Sodium nitride is formed.

Explanation of Solution

For calculating the amount of sodium nitride produced formula is

Number of moles = $\frac{\text{mass}}{\text{molar mass}}$

  $\text{mass = no}\text{. of moles }\times \text{ molar mass}$

No. of moles ${\mathrm{Na}}_{3}N$ = 2moles are formed.

Molar mass =

  $\text{mass = no}\text{. of moles }\times \text{ molar mass}$

  $2moles\times \frac{82.98g}{moles}$

  $\text{mass(g) = 2 }\times \text{ 82}\text{.98g}$

Mass = 165.96g of sodium nitride is formed