Chapter 12.2, Problem 12SP

(a)

Interpretation Introduction

Interpretation:

The number of moles of oxygen required to react with $\text{14}\text{.8 mol Al}$ needs to be calculated.

Concept Introduction :

Mole ratios are the conversion factor by which moles of any reactant or product can be converted into any different mole of reactant or product. The mole ratio is always determined for a balanced chemical reaction.

Answer to Problem 12SP

The moles of oxygen required to react with $\text{14}\text{.8 mol Al}$ is $19.7{\text{ mol O}}_2$

Explanation of Solution

The equation given is:

  $\text{4Al(}s\text{)}+3{\text{O}}_2(g)\to 2{\text{Al}}_{\text{2}}{\text{O}}_3$

The mole ratio conversion is:

  $\frac{1{\text{ mole O}}_2}{\frac{3}{4}\text{ mole Al}}\times 14.8\text{ mol Al = 19}{\text{.7 mol O}}_2$

(b)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{Al}}_2{\text{O}}_3$ formed when $\text{0}{\text{.78 mol O}}_2$ reacts with aluminum needs to be calculated.

Concept Introduction :

Mole ratios are the conversion factor by which moles of any reactant or product can be converted into any different mole of reactant or product. The mole ratio is always determined for a balanced chemical reaction.

Answer to Problem 12SP

The moles of ${\text{Al}}_2{\text{O}}_3$ formed when $\text{0}{\text{.78 mol O}}_2$ reacts with aluminum is $0.52$

Explanation of Solution

The equation given is:

  $\text{4Al(}s\text{)}+3{\text{O}}_2(g)\to 2{\text{Al}}_{\text{2}}{\text{O}}_3$

The mole ratio conversion is:

  $\frac{1{\text{ mole Al}}_2{\text{O}}_3}{\frac{3}{2}{\text{ mole O}}_2}\times 0.78{\text{ mol O}}_2=0.52{\text{ mol Al}}_2{\text{O}}_3$