Chapter 14, Problem 121A

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the reaction given.

Concept Introduction: The representation of the chemical formulas of a substance that react and the substance that are produced is called as chemical equation. In this the number of atoms of the reactants and the atoms of products should be balanced that is the number of atoms should be same on both the sides of a chemical equation.

Answer to Problem 121A

  $\text{4Al}(s)+3O_2(g)\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Explanation of Solution

To write the balanced chemical equation the imbalanced equation that is given is considered first. Now the number of atoms on each side of the reactant arrow is calculated. In this case, in a imbalanced chemical equation there is one aluminum atom and one oxygen molecule on the reactant side. Similarly, there are two aluminum atoms and three oxygen atoms on the product side. Therefore, in order to balance this chemical equation, four moles of aluminum atoms and three moles of oxygen molecules on the reactant side is considered. Similarly, two moles of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is obtained from these reactants which shows that the reaction is balanced.

(b)

Interpretation Introduction

Interpretation: To calculate the mass of each reactant that is required to produce 583 grams of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Concept introduction: The mass of the product that is formed will depend on the moles of the reactant that react to form the product. One mole of the compound will be equal to the molar mass of that compound. The mass of the product that is formed from the reactant will depend on the moles of the reactant that take part in the reaction.

Answer to Problem 121A

617.29 grams of Aluminum and 548.7 grams of oxygen are required.

Explanation of Solution

  $\text{4Al}(s)+3O_2(g)\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}$

From the balanced equation, it is clear that 4 moles of Al are reacting with 3 moles of ${\text{O}}_{\text{2}}$ to produce 2 moles of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

The amount of aluminum required is

  ${\text{583gofAl}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{{\text{2moleofAl}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{102gofAl}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{\text{4moleofAl}}{{\text{2moleofAl}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{\text{27gofAl}}{\text{1moleofAl}}$ =617.29 grams of Al

The amount of oxygen required is

  ${\text{583gofAl}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{{\text{2moleofAl}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{102gofAl}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{{\text{3moleofO}}_2}{{\text{2moleofAl}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{{\text{32gofO}}_2}{{\text{1moleofO}}_2}$ =548.7 grams of oxygen