Chapter 20, Problem 59A

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation of the heating of solid barium chlorate that results in the formation of barium chloride and gaseous oxygen is to be determined.

Concept Introduction: Chemical equations can be balanced by equalizing the atoms present on the reactant and product side.

Answer to Problem 59A

The balanced chemical equation is,

  $Ba{\left(Cl{O}_3\right)}_2\to BaC{l}_2+3O_2$

Explanation of Solution

The heating of solid barium chlorate, $Ba{\left(Cl{O}_3\right)}_2$ results in the formation of barium chloride, $BaC{l}_2$ , and gaseous oxygen. It can be expressed in the chemical equation as,

reaction:

  $Ba{\left(Cl{O}_3\right)}_2\to BaC{l}_2+O_2$

Now, check whether all the elements involved are equal on both sides. The number of oxygen is not equal, so, it can be equalized by multiplying 3 by the $O_2$ . So, the balanced equation becomes,

  $Ba{\left(Cl{O}_3\right)}_2\to BaC{l}_2+3O_2$

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation of solid lead(II) sulfide with gaseous oxygen, which results in the formation of solid lead(II) oxide and gaseous sulfur dioxide is to be determined.

Concept Introduction: Chemical equations can be balanced by equalizing the atoms present on the reactant and product side.

Answer to Problem 59A

The balanced chemical equation is $PbS+3O_2\to 2PbO+2S{O}_2$ .

Explanation of Solution

The reaction of lead(II) sulfide with gaseous oxygen, results in the formation of solid lead(II) oxide and gaseous sulfur dioxide can be written as,

  $PbS+O_2\to PbO+S{O}_2$

The reaction is not in a balanced form since the oxygen count is not equal. It can be made equal by adding 3 to O2, 2 to $PbS$ , 2 to $PbO$ , and 2 to $S{O}_2$ . Therefore, the equation becomes,

  $PbS+3O_2\to 2PbO+2S{O}_2$