Chapter 7, Problem 43A

Interpretation Introduction

Interpretation:

A balanced chemical equation corresponding to the reactions of group I elements with sulfur needs to be presented.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• A balanced chemical equation is one in which the number of atoms of various elements on the reactants side is equal to that on the product side. It depicts the stoichiometry of a given reaction.
• The equations are balanced in two steps:

1) Balance the atoms by matching the number of atoms of each type on the reactants and product side.

2) Adjust the coefficients to the smallest whole number. A coefficient of '1' is implied and therefore not explicitly written.

Answer to Problem 43A

  $\begin{array}{l}\text{2Li(s) + S(s) }\to {\text{ Li}}_2\text{S (s) }\\ \text{2Na(s) + S(s) }\to {\text{ Na}}_2\text{S (s) }\\ \text{2K(s) + S(s) }\to {\text{ K}}_2\text{S (s) }\\ \text{2Rb(s) + S(s) }\to {\text{ Rb}}_2\text{S (s) }\\ \text{2Cs(s) + S(s) }\to {\text{ Cs}}_2\text{S (s) }\\ \text{2Fr(s) + S(s) }\to {\text{ Fr}}_2\text{S (s) }\end{array}$

Explanation of Solution

Group I elements are the alkali metals which include:

Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr)

The required equation involves the reaction between group I metals and solid sulfur to form the corresponding metal sulphide solid. The reactants and products are:

i) Li + S

Reactants:

  $\begin{array}{l}\text{Lithium, solid - Li }(\text{s})\\ \text{Sulfur, solid - S (s)}\end{array}$

Product:

  ${\text{Lithium sulfide, solid - Li}}_2\text{S(s)}$

Unbalanced equation:

  $\text{Li(s) + S(s) }\to {\text{ Li}}_2\text{S (s) }$

The above equation is not balanced since the number of Li atom on the reactants side is 1 while it is 2 on the product side. In order to balance the above equation, multiply Li by 2 to get:

  $\text{2Li(s) + S(s) }\to {\text{ Li}}_2\text{S (s) }$

ii) Na + S

Reactants:

  $\begin{array}{l}\text{Sodium, solid - Na }(\text{s})\\ \text{Sulfur, solid - S (s)}\end{array}$

Product:

  ${\text{Sodium sulfide, solid - Na}}_2\text{S(s)}$

Unbalanced equation:

  $\text{Na(s) + S(s) }\to {\text{ Na}}_2\text{S (s) }$

The above equation is not balanced since the number of Na atom on the reactants side is 1 while it is 2 on the product side. In order to balance the above equation, multiply Na by 2 to get:

  $\text{2Na(s) + S(s) }\to {\text{ Na}}_2\text{S (s) }$

iii) K + S

Reactants:

  $\begin{array}{l}\text{Potassium, solid - K }(\text{s})\\ \text{Sulfur, solid - S (s)}\end{array}$

Product:

  ${\text{Potassium sulfide, solid - K}}_2\text{S(s)}$

Unbalanced equation:

  $\text{K(s) + S(s) }\to {\text{ K}}_2\text{S (s) }$

The above equation is not balanced since the number of K atom on the reactants side is 1 while it is 2 on the product side. In order to balance the above equation, multiply K by 2 to get:

  $\text{2K(s) + S(s) }\to {\text{ K}}_2\text{S (s) }$

iv) Rb + S

Reactants:

  $\begin{array}{l}\text{Rubidium, solid - Rb }(\text{s})\\ \text{Sulfur, solid - S (s)}\end{array}$

Product:

  ${\text{Rubidium sulfide, solid - Rb}}_2\text{S(s)}$

Unbalanced equation:

  $\text{Rb(s) + S(s) }\to {\text{ Rb}}_2\text{S (s) }$

The above equation is not balanced since the number of Rb atom on the reactants side is 1 while it is 2 on the product side. In order to balance the above equation, multiply Rb by 2 to get:

  $\text{2Rb(s) + S(s) }\to {\text{ Rb}}_2\text{S (s) }$

v) Cs + S

Reactants:

  $\begin{array}{l}\text{Cesium, solid - Cs }(\text{s})\\ \text{Sulfur, solid - S (s)}\end{array}$

Product:

  ${\text{Cesium sulfide, solid - Cs}}_2\text{S(s)}$

Unbalanced equation:

  $\text{Cs(s) + S(s) }\to {\text{ Cs}}_2\text{S (s) }$

The above equation is not balanced since the number of Cs atom on the reactants side is 1 while it is 2 on the product side. In order to balance the above equation, multiply Cs by 2 to get:

  $\text{2Cs(s) + S(s) }\to {\text{ Cs}}_2\text{S (s) }$

vi) Fr+ S

Reactants:

  $\begin{array}{l}\text{Francium, solid - Fr }(\text{s})\\ \text{Sulfur, solid - S (s)}\end{array}$

Product:

  ${\text{Francium sulfide, solid - Fr}}_2\text{S(s)}$

Unbalanced equation:

  $\text{Fr(s) + S(s) }\to {\text{ Fr}}_2\text{S (s) }$

The above equation is not balanced since the number of Fr atom on the reactants side is 1 while it is 2 on the product side. In order to balance the above equation, multiply Fr by 2 to get:

  $\text{2Fr(s) + S(s) }\to {\text{ Fr}}_2\text{S (s) }$

Conclusion

The balanced equations corresponding to the reaction of group I metal with sulfur are:

  $\begin{array}{l}\text{2Li(s) + S(s) }\to {\text{ Li}}_2\text{S (s) }\\ \text{2Na(s) + S(s) }\to {\text{ Na}}_2\text{S (s) }\\ \text{2K(s) + S(s) }\to {\text{ K}}_2\text{S (s) }\\ \text{2Rb(s) + S(s) }\to {\text{ Rb}}_2\text{S (s) }\\ \text{2Cs(s) + S(s) }\to {\text{ Cs}}_2\text{S (s) }\\ \text{2Fr(s) + S(s) }\to {\text{ Fr}}_2\text{S (s) }\end{array}$