Chapter 3, Problem 3.139P

(a)

Interpretation Introduction

Interpretation:

The limiting reactant is to be identified.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount(mol) of excess reactant is reactant left after the formation of maximum amount(mol) of products.

The formula to calculate moles is as follows:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$        (1)

The formula to calculate density is,

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$        (2)

(b)

Interpretation Introduction

Interpretation:

The percent yield of the reaction is to be calculated.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The formula to calculate the percentage yield is as follows:

  $\text{Percentage yield}=\left(\frac{\text{actual yield}}{\text{theoretical yield}}\right)100\%$        (4)

(c)

Interpretation Introduction

Interpretation:

The percent atom economy of the reaction is to be calculated.

Concept introduction:

Percent atom economy gives an idea about the efficiency of a synthesis. The formula to calculate the percent atom economy is as follows:

  $\text{\% atom economy}=\left(\frac{\begin{array}{l}\left(\text{Number of moles}\right)\\ \left(\text{Molar mass of desired product}\right)\end{array}}{\begin{array}{l}\text{Sum of}\left(\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)\right)\\ \text{for all product}\end{array}}\right)100\%$