Chapter 3, Problem 3.121P

(a)

Interpretation Introduction

Interpretation:

The mass of ${\text{CO}}_{\text{2}}$ produced upon combustion of $200\text{ g }$ mixture of $25.0\%{\text{ CH}}_{\text{4}}$ and $75.0\%{\text{ C}}_3{\text{H}}_8$ is to be calculated.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

The number of moles is calculated by the formula:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molar mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molar mass g/mol}\right)\left(\text{number of moles}\right)$        (2)

(b)

Interpretation Introduction

Interpretation:

The mass % of ${\text{CH}}_{\text{4}}$ in $252\text{ g}$ of gaseous mixture that contains ${\text{CH}}_{\text{4}}$ and ${\text{C}}_3{\text{H}}_8$ yields $\text{748 g}$ of ${\text{CO}}_{\text{2}}$ is to be determined.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture, multiplied by 100.

The formula to calculate mass % is as follows:

  $\text{Mass }\%=\frac{\text{given mass}\left(\text{g}\right)}{\text{total mass of mixture}\left(\text{g}\right)}\left(100\%\right)$        (5)