Chapter 3, Problem 3.64P

(a)

Interpretation Introduction

Interpretation: Mass of ${\text{O}}_{\text{2}}$ formed when $\text{5}{\text{.50 mL C}}_4{\text{H}}_{\text{10}}$ is burnt completely is to be calculated.

Concept introduction:The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For some stoichiometric balanced equation; the reactant that is present in the smallest amount is consumed completely to form the product and thus governs the product formation.This reactant is known as limiting reactant.On the basis of the stoichiometric ratio by which substrates combine, the moles of product formed from each reactant can be calculated.

(b)

Interpretation Introduction

Interpretation:Moles of ${\text{H}}_{\text{2}}\text{O}$ formed when ${\text{C}}_4{\text{H}}_{\text{10}}$ is burnt completely are to be calculated.

Concept introduction:The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For some stoichiometric balanced equation; the reactant that is present in the smallest amount is consumed completely to form the product and thus governs the product formation.This reactant is known as limiting reactant.On the basis of the stoichiometric ratio by which substrates combine, the moles of product formed from each reactant can be calculated.

(c)

Interpretation Introduction

Interpretation:Total molecules of gases formed are to be calculated.

Concept introduction:The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For some stoichiometric balanced equation; the reactant that is present in the smallest amount is consumed completely to form the product and thus governs the product formation.This reactant is known as limiting reactant.On the basis of the stoichiometric ratio by which substrates combine, the moles of product formed from each reactant can be calculated.