Chapter 5, Problem 5.38E

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 5.38E

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

• $1\text{mol}\text{S}\left(\text{s}\right)+1\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 1\text{mol}{\text{SO}}_2\left(\text{g}\right)$

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

• $\text{32}\text{g}\text{S}+32\text{g}{\text{O}}_{\text{2}}\to 64\text{g}{\text{SO}}_2$

Explanation of Solution

The given reaction is shown below.

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

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.

• $1\text{mol}\text{S}\left(\text{s}\right)+1\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 1\text{mol}{\text{SO}}_2\left(\text{g}\right)$

• 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.

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

• 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.

$\text{32}\text{g}\text{S}+32\text{g}{\text{O}}_{\text{2}}\to 64\text{g}{\text{SO}}_2$

Conclusion

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

• $1\text{mol}\text{S}\left(\text{s}\right)+1\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 1\text{mol}{\text{SO}}_2\left(\text{g}\right)$

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

• $\text{32}\text{g}\text{S}+32\text{g}{\text{O}}_{\text{2}}\to 64\text{g}{\text{SO}}_2$

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 5.38E

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

• $1\text{mol}\text{Sr}\left(\text{s}\right)+2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 1\text{mol}\text{Sr}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+1\text{mol}{\text{H}}_{\text{2}}\left(\text{g}\right)$

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

• $\text{87}\text{.6}\text{g}\text{Sr}+2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}\to 121.6\text{g}\text{Sr}{\left(\text{OH}\right)}_{\text{2}}+2\text{g}{\text{H}}_{\text{2}}$

Explanation of Solution

The given reaction is shown below.

$\text{Sr}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Sr}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

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.

• $1\text{mol}\text{Sr}\left(\text{s}\right)+2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 1\text{mol}\text{Sr}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+1\text{mol}{\text{H}}_{\text{2}}\left(\text{g}\right)$

• 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.

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

• 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.

$\text{87}\text{.6}\text{g}\text{Sr}+2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}\to 121.6\text{g}\text{Sr}{\left(\text{OH}\right)}_{\text{2}}+2\text{g}{\text{H}}_{\text{2}}$

Conclusion

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

• $1\text{mol}\text{Sr}\left(\text{s}\right)+2\text{mol}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 1\text{mol}\text{Sr}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+1\text{mol}{\text{H}}_{\text{2}}\left(\text{g}\right)$

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

• $\text{87}\text{.6}\text{g}\text{Sr}+2\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}\to 121.6\text{g}\text{Sr}{\left(\text{OH}\right)}_{\text{2}}+2\text{g}{\text{H}}_{\text{2}}$

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 5.38E

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

• $2\text{mol}{\text{H}}_2\text{S}\left(\text{g}\right)+3\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2\text{mol}{\text{H}}_2\text{O}\left(\text{g}\right)+2\text{mol}{\text{SO}}_{\text{2}}\left(\text{g}\right)$

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

• $2\left(34\right)\text{g}{\text{H}}_2\text{S}+3\left(32\right)\text{g}{\text{O}}_{\text{2}}\to 2\left(18\right)\text{g}{\text{H}}_2\text{O}+2\left(64\right)\text{g}{\text{SO}}_{\text{2}}$

Explanation of Solution

The given reaction is shown below.

${\text{2H}}_2\text{S}\left(\text{g}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{2H}}_2\text{O}\left(\text{g}\right)+2{\text{SO}}_{\text{2}}\left(\text{g}\right)$

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.

• $2\text{mol}{\text{H}}_2\text{S}\left(\text{g}\right)+3\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2\text{mol}{\text{H}}_2\text{O}\left(\text{g}\right)+2\text{mol}{\text{SO}}_{\text{2}}\left(\text{g}\right)$

• 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.

$\left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{H}}_2\text{S}\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right){\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}2\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}+\\ 2\left(6.022\times {10}^{23}\right){\text{SO}}_{\text{2}}\text{molecules}\end{array}\right)$

• 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.

$2\left(34\right)\text{g}{\text{H}}_2\text{S}+3\left(32\right)\text{g}{\text{O}}_{\text{2}}\to 2\left(18\right)\text{g}{\text{H}}_2\text{O}+2\left(64\right)\text{g}{\text{SO}}_{\text{2}}$

Conclusion

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

• $2\text{mol}{\text{H}}_2\text{S}\left(\text{g}\right)+3\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2\text{mol}{\text{H}}_2\text{O}\left(\text{g}\right)+2\text{mol}{\text{SO}}_{\text{2}}\left(\text{g}\right)$

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

• $2\left(34\right)\text{g}{\text{H}}_2\text{S}+3\left(32\right)\text{g}{\text{O}}_{\text{2}}\to 2\left(18\right)\text{g}{\text{H}}_2\text{O}+2\left(64\right)\text{g}{\text{SO}}_{\text{2}}$

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 5.38E

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

• $\text{4}\text{mol}{\text{NH}}_3\left(\text{g}\right)+5\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 4\text{mol}\text{NO}\left(\text{g}\right)+6\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• $\left(\begin{array}{l}4\left(6.022\times {10}^{23}\right){\text{NH}}_3\text{molecules}+\\ 5\left(6.022\times {10}^{23}\right){\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}4\left(6.022\times {10}^{23}\right)\text{NO}\text{molecules}+\\ 6\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}\end{array}\right)$

• $4\left(17\right)\text{g}{\text{NH}}_3+5\left(32\right)\text{g}{\text{O}}_{\text{2}}\to 4\left(30\right)\text{g}\text{NO}+6\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The given reaction is shown below.

${\text{4NH}}_3\left(\text{g}\right)+5{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{4NO}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

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.

• $\text{4}\text{mol}{\text{NH}}_3\left(\text{g}\right)+5\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 4\text{mol}\text{NO}\left(\text{g}\right)+6\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• 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.

$\left(\begin{array}{l}4\left(6.022\times {10}^{23}\right){\text{NH}}_3\text{molecules}+\\ 5\left(6.022\times {10}^{23}\right){\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}4\left(6.022\times {10}^{23}\right)\text{NO}\text{molecules}+\\ 6\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}\end{array}\right)$

• 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.

$4\left(17\right)\text{g}{\text{NH}}_3+5\left(32\right)\text{g}{\text{O}}_{\text{2}}\to 4\left(30\right)\text{g}\text{NO}+6\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}$

Conclusion

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

• $\text{4}\text{mol}{\text{NH}}_3\left(\text{g}\right)+5\text{mol}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 4\text{mol}\text{NO}\left(\text{g}\right)+6\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• $\left(\begin{array}{l}4\left(6.022\times {10}^{23}\right){\text{NH}}_3\text{molecules}+\\ 5\left(6.022\times {10}^{23}\right){\text{O}}_{\text{2}}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}4\left(6.022\times {10}^{23}\right)\text{NO}\text{molecules}+\\ 6\left(6.022\times {10}^{23}\right){\text{H}}_2\text{O}\text{molecules}\end{array}\right)$

• $4\left(17\right)\text{g}{\text{NH}}_3+5\left(32\right)\text{g}{\text{O}}_{\text{2}}\to 4\left(30\right)\text{g}\text{NO}+6\left(18\right)\text{g}{\text{H}}_{\text{2}}\text{O}$

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 5.38E

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

• $1\text{mol}\text{CaO}\left(\text{s}\right)+3\text{molC}\left(\text{s}\right)\to 1\text{mol}{\text{CaC}}_{\text{2}}\left(\text{s}\right)+1\text{mol}\text{CO}\left(\text{g}\right)$

• $\left(\begin{array}{l}\left(6.022\times {10}^{23}\right)\text{CaO}\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{C}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}\left(6.022\times {10}^{23}\right){\text{CaC}}_{\text{2}}\text{molecules}+\\ \left(6.022\times {10}^{23}\right)\text{CO}\text{molecules}\end{array}\right)$

• $56\text{g}\text{CaO}+3\left(12\right)\text{g}\text{C}\to 64\text{g}{\text{CaC}}_{\text{2}}+28\text{g}\text{CO}$

Explanation of Solution

The given reaction is shown below.

$\text{CaO}\left(\text{s}\right)+3\text{C}\left(\text{s}\right)\to {\text{CaC}}_{\text{2}}\left(\text{s}\right)+\text{CO}\left(\text{g}\right)$

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.

• $1\text{mol}\text{CaO}\left(\text{s}\right)+3\text{molC}\left(\text{s}\right)\to 1\text{mol}{\text{CaC}}_{\text{2}}\left(\text{s}\right)+1\text{mol}\text{CO}\left(\text{g}\right)$

• 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.

$\left(\begin{array}{l}\left(6.022\times {10}^{23}\right)\text{CaO}\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{C}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}\left(6.022\times {10}^{23}\right){\text{CaC}}_{\text{2}}\text{molecules}+\\ \left(6.022\times {10}^{23}\right)\text{CO}\text{molecules}\end{array}\right)$

• 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.

$56\text{g}\text{CaO}+3\left(12\right)\text{g}\text{C}\to 64\text{g}{\text{CaC}}_{\text{2}}+28\text{g}\text{CO}$

Conclusion

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

• $1\text{mol}\text{CaO}\left(\text{s}\right)+3\text{molC}\left(\text{s}\right)\to 1\text{mol}{\text{CaC}}_{\text{2}}\left(\text{s}\right)+1\text{mol}\text{CO}\left(\text{g}\right)$

• $\left(\begin{array}{l}\left(6.022\times {10}^{23}\right)\text{CaO}\text{molecules}+\\ 3\left(6.022\times {10}^{23}\right)\text{C}\text{molecules}\end{array}\right)\to \left(\begin{array}{l}\left(6.022\times {10}^{23}\right){\text{CaC}}_{\text{2}}\text{molecules}+\\ \left(6.022\times {10}^{23}\right)\text{CO}\text{molecules}\end{array}\right)$

• $56\text{g}\text{CaO}+3\left(12\right)\text{g}\text{C}\to 64\text{g}{\text{CaC}}_{\text{2}}+28\text{g}\text{CO}$