Chapter 21, Problem 76GQ

(a)

Interpretation Introduction

Interpretation:

To calculate the mass of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{O}}_{\text{2}}$ produced in the decomposition of ammonium perchlorate.

Concept introduction:

Ammonium perchlorate upon decomposition gives $\text{NO}$, ${\text{Cl}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ gases. It is used as the oxidizer in solid-fuel rockets.

The total number of moles of the substance is equal to the mass of the substance divided by the molecular weight of that substance.

The limiting reagent in the chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by the limiting reagent.

Answer to Problem 76GQ

The mass of water molecule is produced upon decomposition of ammonium perchlorate is $1.94\times 10^{5}kg$.

The mass of oxygen molecule is produced upon decomposition of ammonium perchlorate is $0.864\times 10^{5}kg$.

Explanation of Solution

The mass of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{O}}_{\text{2}}$ produced in the decomposition of ammonium perchlorate is calculated below.

Given:

The mass of ammonium perchlorate is $6.35\times {10}^5\text{kg}$.

The balanced chemical equation of decomposition of ammonium perchlorate is written as,

    $2{\text{NH}}_{\text{4}}{\text{ClO}}_{\text{4}}\left(\text{s}\right)\to 4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)+2\text{NO}\left(\text{g}\right)$                          (1)

The number of moles $\left(n_{{\text{NH}}_{\text{4}}{\text{ClO}}_4}\right)$ of ammonium perchlorate is equal to the mass of ammonium perchlorate divided by its molecular weight. The number of moles $\left(n_{{\text{NH}}_{\text{4}}{\text{ClO}}_4}\right)$ is calculated as,

$\begin{array}{c}{n}_{{\text{NH}}_{\text{4}}{\text{ClO}}_4}=\frac{6.35\times {10}^8\text{g}}{117.49\text{g}\cdot {\text{mol}}^{-1}}\\ =0.05\times {10}^8\text{mol}\end{array}$

From equation (1), two moles of ammonium perchlorate produced four moles of water molecules. Therefore, the number of moles $\left(n_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water molecules produced from $0.05\times {10}^8\text{mol}$ of ammonium perchlorate is written as,

$\begin{array}{c}{n}_{{\text{H}}_{\text{2}}\text{O}}=\left(\frac{4\text{mol}}{2\text{mol}}\right)\left(0.05\times {10}^8\text{mol}\right)\\ =0.108\times {10}^8\text{mol}\end{array}$

The mass $\left(m_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water molecule is equal to the product of a number of moles of the water molecule and its molecular weight. The mass $\left(m_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water molecule is calculated as,

$\begin{array}{c}{m}_{{\text{H}}_{\text{2}}\text{O}}=\left(0.108\times {10}^8\text{mol}\right)\left(18\text{g}\cdot {\text{mol}}^{-1}\right)\\ =1.94\times {10}^8\text{g}\\ =1.94\times {10}^5\text{kg}\end{array}$

Therefore, the mass of water molecule is produced upon decomposition of ammonium perchlorate is $1.94\times {10}^5\text{kg}$.

From equation (1), two moles of ammonium perchlorate produced one mole of the oxygen molecule. Therefore, the number of moles $\left(n_{{\text{O}}_{\text{2}}}\right)$ of oxygen molecule produced from $0.05\times {10}^8\text{mol}$ of ammonium perchlorate is written as,

$\begin{array}{c}{n}_{{\text{O}}_{\text{2}}}=\left(\frac{1\text{mol}}{2\text{mol}}\right)\left(0.05\times {10}^8\text{mol}\right)\\ =0.027\times {10}^8\text{mol}\end{array}$

The mass $\left(m_{{\text{O}}_{\text{2}}}\right)$ of oxygen molecule is equal to the product of a number of moles of the oxygen molecule and its molecular weight. The mass $\left(m_{{\text{O}}_{\text{2}}}\right)$ of oxygen molecule is calculated as,

$\begin{array}{c}{m}_{{\text{O}}_{\text{2}}}=\left(0.027\times {10}^8\text{mol}\right)\left(32\text{g}\cdot {\text{mol}}^{-1}\right)\\ =0.864\times {10}^8\text{g}\\ =0.864\times {10}^5\text{kg}\end{array}$

Therefore, the mass of oxygen molecule is produced upon decomposition of ammonium perchlorate is $0.864\times {10}^5\text{kg}$.

(b)

Interpretation Introduction

Interpretation:

To calculate the mass of aluminium which is required to use up all of the oxygen to form ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Concept introduction:

Ammonium perchlorate upon decomposition gives $\text{NO}$, ${\text{Cl}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ gases. It is used as the oxidizer in solid-fuel rockets.

The total number of moles of the substance is equal to the mass of the substance divided by the molecular weight of that substance.

The limiting reagent in the chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by the limiting reagent.

Answer to Problem 76GQ

The mass of aluminium powder which is required to use up all of the oxygen to form ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is $0.97\times 10^{5}kg$.

Explanation of Solution

The mass of aluminium which is required to use up all of the oxygen to form ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated below.

Given:

The number of moles of oxygen molecule calculated in sub-part (a) is $0.027\times {10}^8\text{mol}$.

The balanced chemical equation of the reaction of aluminium with oxygen molecule is written as,

    $4\text{Al}\left(\text{s}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$ (2)

Since the entire oxygen molecule is used to react with the aluminium powder to produce ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$. Hence, the number of moles of aluminium is equal to the number of moles of the oxygen molecule.

From equation (2), four moles of aluminium reacted with three moles of the oxygen molecule. Therefore, the number of moles $\left(n_{\text{Al}}\right)$ of aluminium is calculated as,

$\begin{array}{c}{n}_{\text{Al}}=\left(\frac{4\text{mol}}{3\text{mol}}\right)\left(0.027\times {10}^8\text{mol}\right)\\ =0.036\times {10}^8\text{mol}\end{array}$

The mass $\left(m_{\text{Al}}\right)$ of aluminium powder is equal to the product of a number of moles of aluminium powder and its molecular weight. The mass $\left(m_{\text{Al}}\right)$ of aluminium powder is calculated as,

$\begin{array}{c}{m}_{\text{Al}}=\left(0.036\times {10}^8\text{mol}\right)\left(26.98\text{g}\cdot {\text{mol}}^{-1}\right)\\ =0.97\times {10}^8\text{g}\\ =0.97\times {10}^5\text{kg}\end{array}$

Therefore, the mass of aluminium powder which is required to use up all of the oxygen to form ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is $0.97\times {10}^5\text{kg}$.

(c)

Interpretation Introduction

Interpretation:

To calculate the mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ produced in the reaction of powdered aluminium with oxygen.

Concept introduction:

Ammonium perchlorate upon decomposition gives $\text{NO}$, ${\text{Cl}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ gases. It is used as the oxidizer in solid-fuel rockets.

The total number of moles of the substance is equal to the mass of the substance divided by the molecular weight of that substance.

The limiting reagent in the chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by the limiting reagent.

Answer to Problem 76GQ

The mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ produced in the reaction of powdered aluminium with oxygen is $1.83\times 10^{5}kg$.

Explanation of Solution

The mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ produced in the reaction of powdered aluminium with oxygen is calculated below.

Given:

The mass of aluminium powder which is required to use up all of the oxygen to form ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ calculated in sub-part (b) is $0.97\times {10}^5\text{kg}$.

The balanced chemical equation of the reaction of aluminium with oxygen molecule is written as,

    $4\text{Al}\left(\text{s}\right)+3{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Four moles of aluminium produced two moles of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ molecule. Therefore, the number of moles $\left(n_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}\right)$ of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ molecule is calculated as,

$\begin{array}{c}{n}_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}=\left(\frac{2\text{mol}}{4\text{mol}}\right)\left(0.036\times {10}^8\text{mol}\right)\\ =0.018\times {10}^8\text{mol}\end{array}$

The mass $\left(m_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}\right)$ of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ molecule is equal to the product of a number of moles of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ molecule and its molecular weight. The mass $\left(m_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}\right)$ of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ molecule is calculated as,

$\begin{array}{c}{m}_{\text{Al}}=\left(0.018\times {10}^8\text{mol}\right)\left(101.96\text{g}\cdot {\text{mol}}^{-1}\right)\\ =1.83\times {10}^8\text{g}\\ =1.83\times {10}^5\text{kg}\end{array}$

Therefore, the mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ produced in the reaction of powdered aluminium with oxygen is $1.83\times {10}^5\text{kg}$.