Chapter 4, Problem 38P

4-54 Answer true or false.

(a) Stoichiometry is the study of mass relationships in chemical reactions.

(b) To determine mass relationships in a chemical reaction, you first need to know the balanced chemical equation for the reaction.

(c) To convert from grams to moles and vice versa, use Avogadro’s number as a conversion factor.

(d) To convert from grams to moles and vice versa, use molar mass as a conversion factor.

(e) A limiting reagent is the reagent that is used up first.

(f) Suppose a chemical reaction between A and B requires 1 mol of A and 2 mol of B. If 1 mol of each is present, then B is the limiting reagent.

(g) Theoretical yield is the yield of product that should be obtained according to the balanced chemical equation.

(h) Theoretical yield is the yield of product that should be obtained if all limiting reagent is con verted to product.

(j) Percent yield is the number of grams of product divided by the number of grams of the limiting reagent times 100.

(j) To calculate percent yield, divide the mass of product formed by the theoretical yield and multiply by 100.

Interpretation Introduction

(a)

Interpretation:

State true or false for the following statement: Stoichiometry is the study of mass relationships in chemical reactions.

Concept Introduction:

Stoichiometry is a quantitative relationship between number of reactants and products in a balanced chemical equation.

Answer to Problem 38P

Stoichiometry is the study of mass relationships in chemical reactions.

Thus the statement is true.

Explanation of Solution

Reason for true statement:

Stoichiometry is the quantitative relationship between the amount of reactants consumed and the amount of product formed in a balanced chemical equation.

Since, stoichiometry deals with the amount of the reactants and products we can say that stoichiometry is the study of mass relationship in chemical reactions.

Interpretation Introduction

(b)

Interpretation:

State true or false for the following statement: To determine mass relationships in a chemical reaction, you first need to know the balanced chemical equation for the reaction.

Concept Introduction:

Stoichiometry is a quantitative relationship between number of reactants and products in a balanced chemical equation.

Answer to Problem 38P

To determine mass relationships in a chemical reaction, you first need to know the balanced chemical equation for the reaction.

Thus, the statement is true.

Explanation of Solution

Reason for true statement:

In a balanced chemical equation the coefficients represent the number of molecules and as the number of molecules are proportional to the number of moles therefore the coefficients in a balanced chemical equation also represent number of moles.

In a balanced chemical equation the number of moles is represented by the coefficients present in them and by knowing the number the number of moles we can determine mass relationships in a chemical reaction.

Interpretation Introduction

(c)

Interpretation:

State true or false for the following statement: To convert from grams to mole and vice versa, use Avogadro’s number as a coversion factor.

Concept Introduction:

Mole:

One mole of a substance is the amount that contains as many atoms, molecules or ions as there are atoms in exactly 12 g of carbon-12.

Molar mass:

Molar mass of any substance is the formula weight of the substance expressed in grams per mole.

Answer to Problem 38P

To convert from grams to mole and vice versa, use molar mass as a coversion factor.

Thus the statement is false.

Explanation of Solution

Reason for false statement:

Molar mass is used as the conversion factor to convert grams into moles or vice-versa.

Since molar mass is the mass of one mole substance therefore it can be used to convert grams to moles or moles to grams as follows:

Suppose if we want to convert 54 g of water into moles. Then,

We know that molar mass of water is 18.0 g/mol. Therefore,

$\begin{array}{l}Numb\text{e}rofm\text{O}\text{l}\text{e}sofw\text{a}t\text{e}rin54g=54\sout{g{\text{H}}_2\text{O}}\times \frac{1m\text{O}\text{l}{\text{H}}_2\text{O}}{18\sout{g{\text{H}}_2\text{O}}}\\ =3m\text{O}\text{l}{\text{H}}_2\text{O}\end{array}$

Similarly if we want to convert 4 moles of water into grams. Then,

We know that molar mass of water is 18.0 g/mol. Therefore,

$\begin{array}{l}Numb\text{e}rofgr\text{a}msofw\text{a}t\text{e}rin4m\text{O}\text{l}=4\sout{m\text{O}\text{l}{\text{H}}_2\text{O}}\times \frac{18g{\text{H}}_2\text{O}}{1\sout{m\text{O}\text{l}{\text{H}}_2\text{O}}}\\ =72g{\text{H}}_2\text{O}\end{array}$

Molar mass is used as the conversion factor to convert grams into moles or vice-versa.

Interpretation Introduction

(d)

Interpretation:

State true or false for the following statement: To convert from grams to moles and vice versa, use molar mass as conversion factor.

Mole:

One mole of a substance is the amount that contains as many atoms, molecules or ions as there are atoms in exactly 12 g of carbon-12.

Molar mass:

Molar mass of any substance is the formula weight of the substance expressed in grams per mole.

Answer to Problem 38P

To convert from grams to moles and vice versa, use molar mass as conversion factor.

Thus the statement is true.

Explanation of Solution

Reason for true statement:

Molar mass is used as the conversion factor to convert grams into moles or vice-versa.

Since molar mass is the mass of one mole substance therefore it can be used to convert grams to moles or moles to grams as follows:

Suppose if we want to convert 54 g of water into moles. Then,

We know that molar mass of water is 18.0 g/mol. Therefore,

$\begin{array}{l}Numb\text{e}rofm\text{O}\text{l}\text{e}sofw\text{a}t\text{e}rin54g=54\sout{g{\text{H}}_2\text{O}}\times \frac{1m\text{O}\text{l}{\text{H}}_2\text{O}}{18\sout{g{\text{H}}_2\text{O}}}\\ =3m\text{O}\text{l}{\text{H}}_2\text{O}\end{array}$

Similarly if we want to convert 4 moles of water into grams. Then,

We know that molar mass of water is 18.0 g/mol. Therefore,

$\begin{array}{l}Numb\text{e}rofgr\text{a}msofw\text{a}t\text{e}rin4m\text{O}\text{l}=4\sout{m\text{O}\text{l}{\text{H}}_2\text{O}}\times \frac{18g{\text{H}}_2\text{O}}{1\sout{m\text{O}\text{l}{\text{H}}_2\text{O}}}\\ =72g{\text{H}}_2\text{O}\end{array}$.

Interpretation Introduction

(e)

Interpretation:

State true or false for the following statement: A limiting reagent is the reagent that is used up first.

Concept Introduction:

The reactant which used up first leaving the excess reagent in a chemical reaction is known as limiting reagent.

Answer to Problem 38P

A limiting reagent is the reagent that is used up first.

Thus, the statement is true.

Explanation of Solution

Reason for true statement:

Limiting reagent is the reactant which is used up first in a chemical reaction.

A limiting reagent is the reactant which is used up first. Consider the following reaction.

$\begin{array}{l}{\text{N}}_2(g)+{\text{O}}_2(g)\underset{}{\overset{}{\to }}2\text{N}\text{O}(g)\\ b\text{e}\text{F}\text{O}r\text{e}r\text{e}\text{a}\text{C}t\text{i}\text{O}n(inm\text{O}\text{l}\text{e}s)10.04.00\\ \text{a}\text{F}t\text{e}rr\text{e}\text{a}\text{C}t\text{i}\text{O}n(inm\text{O}\text{l}\text{e}s)6.004.0\end{array}$

In this reaction 4 moles of oxygen react with only 4 moles of nitrogen. At the end of the reaction oxygen is completely used up and 6 moles of nitrogen remains unreacted. Therefore, in this reaction oxygen is limiting reagent.

Interpretation Introduction

(f)

Interpretation:

State true or false for the following statement: Suppose a chemical reaction between A and B requires 1 mol of A and 2 mol of B. If 1 mol of each is present, then B is the limiting reagent.

Concept Introduction:

The reactant which used up first leaving the excess reagent in a chemical reaction is known as limiting reagent.

Answer to Problem 38P

Suppose a chemical reaction between A and B requires 1 mol of A and 2 mol of B. If 1 mol of each is present, then B is the limiting reagent.

Thus the statement is true.

Explanation of Solution

Reason for true statement:

Limiting reagent is the reactant which is used up first in a chemical reaction.

A limiting reagent is the reactant which is used up first. Consider in the following reaction.

$\begin{array}{l}\text{a}+2B\underset{}{\overset{}{\to }}pr\text{O}du\text{C}t\\ b\text{e}\text{F}\text{O}r\text{e}r\text{e}\text{a}\text{C}t\text{i}\text{O}n(inm\text{O}\text{l}\text{e}s)1.01.00\\ \text{a}\text{F}t\text{e}rr\text{e}\text{a}\text{C}t\text{i}\text{O}n(inm\text{O}\text{l}\text{e}s)0.500.5\end{array}$

In this reaction 1 moles of reactant B reacts with only 0.5 mole of reactant A. At the end of the reaction reactant B is completely used up and 0.5 mole of reactant remains unreacted. Therefore, in this reaction reactant B is limiting reagent.

Interpretation Introduction

(g)

Interpretation:

State true or false for the following statement: Theoretical yield is the yield of product that should be obtained according to the balanced chemical equation.

Concept Introduction:

Theoretical yield is the maximum expected amount of product that should be form in a chemical reaction according to the stoichiometry of the balanced equation.

Answer to Problem 38P

Theoretical yield is the yield of product that should be obtained according to the balanced chemical equation.

Thus, the statement is true.

Explanation of Solution

Reason for true statement:

Theoretical yield is the yield of the product according to the balanced chemical equation.

Theoretical yield is the mass of the product form in a chemical reaction according to the stoichiometry of the balanced equation.

Interpretation Introduction

(h)

Interpretation:

State true or false for the following statement: Theoretical yield is the yield of product that should be obtained if all limiting reagent is converted to product.

Concept Introduction:

Theoretical yield is the maximum expected amount of product that should be form in a chemical reaction according to the stoichiometry of the balanced equation.

Answer to Problem 38P

Theoretical yield is the yield of product that should be obtained if all limiting reagent is converted to product.

Thus, the statement is true.

Explanation of Solution

Reason for true statement:

Theoretical yield is the mass of the product form in a chemical reaction according to the stoichiometry of the balanced equation.

Limiting reagent is the reactant that is used up first in the chemical reaction and theoretical yield is the yield of product that should be form in a chemical reaction according to the stoichiometry of the balanced equation. In a balanced chemical equation, the amount of product depends upon the limiting reagent.

Interpretation Introduction

(i)

Interpretation:

State true or false for the following statement: Percent yield is the number of grams of product divided by the number of grams of the limiting reagent times 100.

Concept Introduction:

Percentage yield is the percent ratio of actual yield to the theoretical yield.

Answer to Problem 38P

Percent yield is the number of grams of product divided by the number of grams of the limiting reagent times 100.

Thus, the statement is true.

Explanation of Solution

Reason for true statement:

Percentage yield is the actual yield divided by theoretical yield into hundred.

In a balanced chemical equation theoretical yield depends upon the limiting reagent therefore it can be concluded that theoretical yield depends upon the number of gram of the limiting reagent. Therefore, percent yield is the number of grams of product divided by the number of grams of the limiting reagent times 100.

Interpretation Introduction

(j)

Interpretation:

State true or false for the following statement: To calculate percent yield, divide the mass of product formed by the theoretical yield and multiply by 100.

Concept Introduction:

Actual yield is the actual amount formed in a chemical reaction.

Theoretical yield is the maximum expected amount of product that should be form in a chemical reaction according to the stoichiometry of the balanced equation.

Percentage yield is the percent ratio of actual yield to the theoretical yield.

Answer to Problem 38P

To calculate percent yield, divide the mass of product formed by the theoretical yield and multiply by 100.

Thus, the statement is true.

Explanation of Solution

Reason for true statement:

Actual yield is the actual amount of product formed.

Actual yield is the mass of the product formed and we know that percentage yield is the percent ratio of the actual yield and theoretical yield.