Chapter 11, Problem 65A

(a)

Interpretation Introduction

Interpretation:

The balance chemical equation for the reaction of solution of potassium chromate with solution of lead (II) nitrate needs to be determined. It produces a yellow precipitate of lead (II) chromate with a solution of potassium nitrate.

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 65A

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}{\text{ + 2 KNO}}_{\text{3(aq)}}$

Explanation of Solution

The reaction of solution of potassium chromate with solution of lead (II) nitrate. It produces a yellow precipitate of lead (II) chromate with a solution of potassium nitrate.

The balance reaction must have all the atoms same at both sides of the reaction equation. Thus the balance equation can be written as:

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}{\text{ + 2 KNO}}_{\text{3(aq)}}$

(b)

Interpretation Introduction

Interpretation:

The mass of lead chromate formed from 0.250 moles of potassium chromate for the given reaction needs to be determined.

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}{\text{ + 2 KNO}}_{\text{3(aq)}}$

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 65A

$\text{80}{\text{.7 g PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}$

Explanation of Solution

The reaction of solution of potassium chromate with solution of lead (II) nitrate. It produces a yellow precipitate of lead (II) chromate with a solution of potassium nitrate.

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}\stackrel{}{\to }{\text{ PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}{\text{ + 2 KNO}}_{\text{3(aq)}}$

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation.

Moles of potassium chromate = 0.250 moles

Molar mass of lead chromate = 323 g/mol

1 mole of sodium potassium chromate = 1 mole of lead chromate

Calculate mass of lead chromate with the help of mole ratio:

$\text{0}{\text{.250 moles K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}\text{ ×}\frac{{\text{1 mole PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}}{{\text{1 moles K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7(aq)}}}\text{×}\frac{{\text{ 323 g PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}}{{\text{1 mole PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}}\text{= 80}{\text{.7 g PbCr}}_{\text{2}}{\text{O}}_{\text{7(s)}}$