Chapter 4, Problem 4.151P

(a)

Interpretation Introduction

Interpretation:

The complete combustion of ethanol and of gasoline is to be written.

Concept introduction:

A combustion reaction is a reaction in which reactant is reacted with molecular oxygen to form the product. Heat is released and the energy is produced in the reaction. Molecular oxygen is employed as an oxidizing agent in these reactions.

(b)

Interpretation Introduction

Interpretation:

The mass of oxygen required to burn completely $1.00\text{ L}$ of a mixture is to be determined.

Concept introduction:

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is ${\text{kg/m}}^{\text{3}}$. The formula to calculate density is,

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

(c)

Interpretation Introduction

Interpretation:

The volume of ${\text{O}}_2$ needed to burn $1.00\text{ L}$ of the mixture is to be calculated.

Concept introduction:

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

  $\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

(d)

Interpretation Introduction

Interpretation:

The volume of air that is needed to burn $1.00\text{L}$ of the mixture is to be determined.

Concept introduction:

Volume percent is defined as the volume of a component divided by the total volume of the mixture multiplied by 100.

The formula to calculate the volume percent is as follows:

  $\text{Volume}\text{percent}=\left(\frac{\text{Volume}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}\right)100$