Chapter 9, Problem 11A

a.

Interpretation Introduction

Interpretation: The number of moles of $\text{KCl}$ producedwhen 3.5 mole of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ are produced in the given reaction needs to be calculated.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 11A

  $\text{5}\text{.83 mol of KCl}$ .

Explanation of Solution

Given:

  ${\text{3P}}_4(s)+{\text{ 10KClO}}_3(s)\stackrel{\text{heat}}{\to }\text{ 10KCl}(s){\text{ + 6P}}_{\text{2}}{\text{O}}_{\text{5}}\text{(}s\text{)}$

From the balanced chemical reaction, the mole ratio of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ and $\text{KCl}$ is 6:10 so, the number of moles of $\text{KCl}$ produced when 3.5 moles of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ is produced are:

  $\text{3}{\text{.5 mol of P}}_{\text{2}}{\text{O}}_{\text{5}}\text{×}\frac{\text{10 mol of KCl}}{{\text{6 mol of P}}_{\text{2}}{\text{O}}_{\text{5}}}\text{ = 5}\text{.83 mol of KCl}$

Hence, number of moles of $\text{KCl}$ produced when 3.5 mole of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ are produced in the given reaction is $\text{5}\text{.83 mol}$ .

b.

Interpretation Introduction

Interpretation: The number of moles of ${\text{P}}_{\text{4}}$ required when 3.5 mole of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ are produced in the given reaction needs to be calculated.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Answer to Problem 11A

  $\text{1}{\text{.75 mol of P}}_4$ .

Explanation of Solution

The balanced reaction is:

  ${\text{3P}}_4(s)+{\text{ 10KClO}}_3(s)\stackrel{\text{heat}}{\to }\text{ 10KCl}(s){\text{ + 6P}}_{\text{2}}{\text{O}}_{\text{5}}\text{(}s\text{)}$

From the balanced chemical reaction, the mole ratio of ${\text{P}}_4$ and ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ is 3:6 so, the number of moles of ${\text{P}}_4$ required when 3.5 moles of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ is produced are:

  $\text{3}{\text{.5 mol of P}}_{\text{2}}{\text{O}}_{\text{5}}\text{×}\frac{{\text{3 mol of P}}_4}{{\text{6 mol of P}}_{\text{2}}{\text{O}}_{\text{5}}}\text{ = 1}{\text{.75 mol of P}}_4$

Hence, number of moles of ${\text{P}}_{\text{4}}$ required when 3.5 mole of ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ are produced in the given reactionis $\text{1}\text{.75 mol}$ .