Chapter 5, Problem 5.39E

Interpretation Introduction

(a)

Interpretation:

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are to be stated.

Concept introduction:

Stoichiometry of a chemical species involved in a chemical reaction represents the amount of chemical species involved in the chemical reaction. The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product. Stoichiometry of a chemical species is also represented in number of moles. The number of moles of a substance is given as,

$n=\frac{m}{M}$

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Answer to Problem 5.39E

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• ${\text{2SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{SO}}_3\left(\text{g}\right)$

• $\left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{SO}}_2\text{molecules}+\\ 6.022\times {10}^{23}{\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to 2\left(6.022\times {10}^{23}\right){\text{SO}}_3\text{molecules}$

• $\text{2}\left(64\right)\text{g}{\text{SO}}_2+32\text{g}{\text{O}}_{\text{2}}\to 2\left(80\right)\text{g}{\text{SO}}_3$

Explanation of Solution

The given reaction is shown below.

${\text{2SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{SO}}_3\left(\text{g}\right)$

The statements for the above reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• The coefficients of the reactants and products can be written as the number of the moles of reactants and products. The statement for the given reaction is represented as follows.

${\text{2SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{SO}}_3\left(\text{g}\right)$

• There are $6.022\times {10}^{23}$ particles present in $1\text{mol}$ of any substance. The statement for the given reaction is represented as follows.

$\left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{SO}}_2\text{molecules}+\\ 6.022\times {10}^{23}{\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to 2\left(6.022\times {10}^{23}\right){\text{SO}}_3\text{molecules}$

• The molecular weight of the compound is equal to the one mole of a compound. The statement for the given reaction is represented as follows.

$\text{2}\left(64\right)\text{g}{\text{SO}}_2+32\text{g}{\text{O}}_{\text{2}}\to 2\left(80\right)\text{g}{\text{SO}}_3$

Conclusion

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• ${\text{2SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{SO}}_3\left(\text{g}\right)$

• $\left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{SO}}_2\text{molecules}+\\ 6.022\times {10}^{23}{\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to 2\left(6.022\times {10}^{23}\right){\text{SO}}_3\text{molecules}$

• $\text{2}\left(64\right)\text{g}{\text{SO}}_2+32\text{g}{\text{O}}_{\text{2}}\to 2\left(80\right)\text{g}{\text{SO}}_3$

Interpretation Introduction

(b)

Interpretation:

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are to be stated.

Concept introduction:

Stoichiometry of a chemical species involved in a chemical reaction represents the amount of chemical species involved in the chemical reaction. The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product. Stoichiometry of a chemical species is also represented in number of moles. The number of moles of a substance is given as,

$n=\frac{m}{M}$

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Answer to Problem 5.39E

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $\text{4}\text{mol}\text{HCl}\left(\text{g}\right)+1\text{mol}{\text{O}}_2\left(\text{g}\right)\to 2\text{mol}{\text{Cl}}_2\left(\text{g}\right)+2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• $\left(\begin{array}{l}4\left(6.022\times {10}^{23}\right)\text{HCl}\text{molecules}+\\ \left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{Cl}}_2\text{molecules}+\\ 2\left(6.022\times {10}^{23}\right){\text{H}}_{\text{2}}\text{O}\text{molecules}\end{array}\right)$

• $4\left(36.5\right)\text{g}\text{HCl}+1\left(16\right)\text{g}{\text{O}}_2\to 2\left(71\right)\text{g}{\text{Cl}}_2$

(g)+2(18) g H2O

Explanation of Solution

The given reaction is shown below.

$\text{4HCl}\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\to {\text{2Cl}}_2\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The statements for the above reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• The coefficients of the reactants and products can be written as the number of the moles of reactants and products. The statement for the given reaction is represented as follows.

$\text{4}\text{mol}\text{HCl}\left(\text{g}\right)+1\text{mol}{\text{O}}_2\left(\text{g}\right)\to 2\text{mol}{\text{Cl}}_2\left(\text{g}\right)+2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• There are $6.022\times {10}^{23}$ particles present in $1\text{mol}$ of any substance. The statement for the given reaction is represented as follows.

$\left(\begin{array}{l}4\left(6.022\times {10}^{23}\right)\text{HCl}\text{molecules}+\\ \left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{Cl}}_2\text{molecules}+\\ 2\left(6.022\times {10}^{23}\right){\text{H}}_{\text{2}}\text{O}\text{molecules}\end{array}\right)$

• The molecular weight of the compound is equal to the one mole of a compound. The statement for the given reaction is represented as follows.

$4\left(36.5\right)\text{g}\text{HCl}+1\left(16\right)\text{g}{\text{O}}_2\to 2\left(71\right)\text{g}{\text{Cl}}_2\left(\text{g}\right)+2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}$

Conclusion

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $\text{4}\text{mol}\text{HCl}\left(\text{g}\right)+1\text{mol}{\text{O}}_2\left(\text{g}\right)\to 2\text{mol}{\text{Cl}}_2\left(\text{g}\right)+2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• $\left(\begin{array}{l}4\left(6.022\times {10}^{23}\right)\text{HCl}\text{molecules}+\\ \left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{Cl}}_2\text{molecules}+\\ 2\left(6.022\times {10}^{23}\right){\text{H}}_{\text{2}}\text{O}\text{molecules}\end{array}\right)$

• $4\left(36.5\right)\text{g}\text{HCl}+1\left(16\right)\text{g}{\text{O}}_2\to 2\left(71\right)\text{g}{\text{Cl}}_2\left(\text{g}\right)+2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}$

Interpretation Introduction

(c)

Interpretation:

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are to be stated.

Concept introduction:

Stoichiometry of a chemical species involved in a chemical reaction represents the amount of chemical species involved in the chemical reaction. The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product. Stoichiometry of a chemical species is also represented in number of moles. The number of moles of a substance is given as,

$n=\frac{m}{M}$

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Answer to Problem 5.39E

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $1\text{mol}{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)+3\text{C}\left(\text{s}\right)\to 2\text{mol}\text{Fe}\left(\text{s}\right)+3\text{mol}\text{CO}\left(\text{g}\right)$

• $\left(\begin{array}{l}\left(6.022\times {10}^{23}\right){\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{C}\text{atoms}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right)\text{Fe}\text{atoms}+\\ 3\left(6.022\times {10}^{23}\right)\text{CO}\text{molecules}\end{array}\right)$

• $159.6\text{g}{\text{Fe}}_2{\text{O}}_3+3\left(12\right)\text{g}\text{C}\to 2\left(55.8\right)\text{g}\text{Fe}+3\left(28\right)\text{g}\text{CO}$

Explanation of Solution

The given reaction is shown below.

${\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)+3\text{C}\left(\text{s}\right)\to \text{2Fe}\left(\text{s}\right)+3\text{CO}\left(\text{g}\right)$

The statements for the above reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• The coefficients of the reactants and products can be written as the number of the moles of reactants and products. The statement for the given reaction is represented as follows.

$1\text{mol}{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)+3\text{C}\left(\text{s}\right)\to 2\text{mol}\text{Fe}\left(\text{s}\right)+3\text{mol}\text{CO}\left(\text{g}\right)$

• There are $6.022\times {10}^{23}$ particles present in $1\text{mol}$ of any substance. The statement for the given reaction is represented as follows.

$\left(\begin{array}{l}\left(6.022\times {10}^{23}\right){\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{C}\text{atoms}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right)\text{Fe}\text{atoms}+\\ 3\left(6.022\times {10}^{23}\right)\text{CO}\text{molecules}\end{array}\right)$

• The molecular weight of the compound is equal to the one mole of a compound. The statement for the given reaction is represented as follows.

$159.6\text{g}{\text{Fe}}_2{\text{O}}_3+3\left(12\right)\text{g}\text{C}\to 2\left(55.8\right)\text{g}\text{Fe}+3\left(28\right)\text{g}\text{CO}$

Conclusion

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $1\text{mol}{\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)+3\text{C}\left(\text{s}\right)\to 2\text{mol}\text{Fe}\left(\text{s}\right)+3\text{mol}\text{CO}\left(\text{g}\right)$

• $\left(\begin{array}{l}\left(6.022\times {10}^{23}\right){\text{Fe}}_2{\text{O}}_3\left(\text{s}\right)\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{C}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right)\text{Fe}\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{CO}\text{molecules}\end{array}\right)$

• $159.6\text{g}{\text{Fe}}_2{\text{O}}_3+3\left(12\right)\text{g}\text{C}\to 2\left(55.8\right)\text{g}\text{Fe}+3\left(28\right)\text{g}\text{CO}$

Interpretation Introduction

(d)

Interpretation:

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are to be stated.

Concept introduction:

Stoichiometry of a chemical species involved in a chemical reaction represents the amount of chemical species involved in the chemical reaction. The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product. Stoichiometry of a chemical species is also represented in number of moles. The number of moles of a substance is given as,

$n=\frac{m}{M}$

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Answer to Problem 5.39E

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $2\text{mol}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+\to 2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(l\right)+\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)$

• $\left(2\left(6.022\times {10}^{23}\right){\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{molecules}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}+\\ \left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)$

• $2\left(34\right)\text{g}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\to 2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}+\left(32\right)\text{g}{\text{O}}_2$

Explanation of Solution

The given reaction is shown below.

${\text{2H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{O}}_2\left(\text{g}\right)$

The statements for the above reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• The coefficients of the reactants and products can be written as the number of the moles of reactants and products. The statement for the given reaction is represented as follows.

$2\text{mol}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+\to 2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(l\right)+\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)$

• There are $6.022\times {10}^{23}$ particles present in $1\text{mol}$ of any substance. The statement for the given reaction is represented as follows.

$\left(2\left(6.022\times {10}^{23}\right){\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{molecules}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}+\\ \left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)$

• The molecular weight of the compound is equal to the one mole of a compound. The statement for the given reaction is represented as follows.

$2\left(34\right)\text{g}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\to 2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}+\left(32\right)\text{g}{\text{O}}_2$

Conclusion

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $2\text{mol}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+\to 2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(l\right)+\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)$

• $\left(2\left(6.022\times {10}^{23}\right){\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{molecules}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}+\\ \left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)$

• $2\left(34\right)\text{g}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\to 2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}+\left(32\right)\text{g}{\text{O}}_2$

Interpretation Introduction

(e)

Interpretation:

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are to be stated.

Concept introduction:

Stoichiometry of a chemical species involved in a chemical reaction represents the amount of chemical species involved in the chemical reaction. The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product. Stoichiometry of a chemical species is also represented in number of moles. The number of moles of a substance is given as,

$n=\frac{m}{M}$

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Answer to Problem 5.39E

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $2\text{mol}\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}\left(\text{g}\right)+9\text{mol}{\text{O}}_2\left(\text{g}\right)\to 6\text{mol}{\text{CO}}_{\text{2}}\left(\text{g}\right)+6\text{mol}{\text{H}}_2\text{O}\left(\text{g}\right)$

• $\left(\begin{array}{c}2\left(6.022\times {10}^{23}\right)\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}\text{molecules}+\\ 9\left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)\to \left(\begin{array}{l}6\left(6.022\times {10}^{23}\right){\text{CO}}_{\text{2}}\text{molecules}+\\ 6\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}\end{array}\right)$

• $2\left(42\right)\text{g}\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}+9\left(32\right)\text{g}{\text{O}}_2\to 6\left(44\right)\text{g}{\text{CO}}_{\text{2}}+6\left(18\right)\text{g}{\text{H}}_2\text{O}$

Explanation of Solution

The given reaction is shown below.

$\text{2C}{\text{3}}_{}{\text{H}}_{\text{6}}\left(\text{g}\right)+9{\text{O}}_2\left(\text{g}\right)\to 6{\text{CO}}_{\text{2}}\left(\text{g}\right)+6{\text{H}}_2\text{O}\left(\text{g}\right)$

The statements for the above reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• The coefficients of the reactants and products can be written as the number of the moles of reactants and products. The statement for the given reaction is represented as follows.

$2\text{mol}\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}\left(\text{g}\right)+9\text{mol}{\text{O}}_2\left(\text{g}\right)\to 6\text{mol}{\text{CO}}_{\text{2}}\left(\text{g}\right)+6\text{mol}{\text{H}}_2\text{O}\left(\text{g}\right)$

• There are $6.022\times {10}^{23}$ particles present in $1\text{mol}$ of any substance. The statement for the given reaction is represented as follows.

$\left(\begin{array}{c}2\left(6.022\times {10}^{23}\right)\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}\text{molecules}+\\ 9\left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)\to \left(\begin{array}{l}6\left(6.022\times {10}^{23}\right){\text{CO}}_{\text{2}}\text{molecules}+\\ 6\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}\end{array}\right)$

• The molecular weight of the compound is equal to the one mole of a compound. The statement for the given reaction is represented as follows.

$2\left(42\right)\text{g}\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}+9\left(32\right)\text{g}{\text{O}}_2\to 6\left(44\right)\text{g}{\text{CO}}_{\text{2}}+6\left(18\right)\text{g}{\text{H}}_2\text{O}$

Conclusion

The statements for the given reaction equivalent to Statements $2$, $3$, and $4$ given in Section 5.9 are,

• $2\text{mol}\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}\left(\text{g}\right)+9\text{mol}{\text{O}}_2\left(\text{g}\right)\to 6\text{mol}{\text{CO}}_{\text{2}}\left(\text{g}\right)+6\text{mol}{\text{H}}_2\text{O}\left(\text{g}\right)$

• $\left(\begin{array}{c}2\left(6.022\times {10}^{23}\right)\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}\text{molecules}+\\ 9\left(6.022\times {10}^{23}\right){\text{O}}_2\text{molecules}\end{array}\right)\to \left(\begin{array}{l}6\left(6.022\times {10}^{23}\right){\text{CO}}_{\text{2}}\text{molecules}+\\ 6\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}\end{array}\right)$

• $2\left(42\right)\text{g}\text{C}{\text{3}}_{}{\text{H}}_{\text{6}}+9\left(32\right)\text{g}{\text{O}}_2\to 6\left(44\right)\text{g}{\text{CO}}_{\text{2}}+6\left(18\right)\text{g}{\text{H}}_2\text{O}$