Chapter 11.1, Problem 1PP

(a)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. It has to be interpreted in terms of particles, moles and mass and obeying of law of conservation of mass in it has to be shown.

${\text{N}}_{\text{2}}\left(g\right)+3{\text{H}}_2\left(g\right)\to {\text{2NH}}_{\text{3}}\left(g\right)$

Concept introduction:

According to law of conservation of mass, it is not possible to form the mass or to destroy the mass. The mass always remains conserved.

The number of moles is calculated as shown below:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

Answer to Problem 1PP

The given equation shows that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to form two molecules of ammonia gas. In terms of number of moles it is seen that one mole of nitrogen gas reacts with three moles of hydrogen gas to form two moles of ammonia gas. Here, $28.0\text{g}$ nitrogen reacts with $6.0\text{g}$ of hydrogen to form $34.0\text{g}$ of ammonia. The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below:

${\text{N}}_{\text{2}}\left(g\right)+3{\text{H}}_2\left(g\right)\to {\text{2NH}}_{\text{3}}\left(g\right)$

The above equation shows that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to form two molecules of ammonia gas. In terms of number of moles, it is seen that one mole of nitrogen gas reacts with three moles of hydrogen gas to form two moles of ammonia gas.

The molar mass of nitrogen is $\text{28}\text{.0}\text{g/mol}$. The mass of nitrogen is calculated as shown below:

$\begin{array}{c}\text{1}\text{mol}=\frac{\text{Given mass}}{\text{28}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 28.\text{0}\frac{\text{g}}{\text{mol}}\\ =28.0\text{g}\end{array}$

The molar mass of hydrogen is $\text{2}\text{.0}\text{g/mol}$. The mass of hydrogen is calculated as shown below:

$\begin{array}{c}3\text{mol}=\frac{\text{Given mass}}{\text{2}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=3\text{mol}\times 2.\text{0}\frac{\text{g}}{\text{mol}}\\ =6.0\text{g}\end{array}$

The molar mass of ammonia is $\text{17}\text{.0}\text{g/mol}$. The mass of ammonia is calculated as shown below:

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{\text{17}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=2\text{mol}\times 17.\text{0}\frac{\text{g}}{\text{mol}}\\ =34.0\text{g}\end{array}$

Thus, $28.0\text{g}$ nitrogen reacts with $6.0\text{g}$ of hydrogen to form $34.0\text{g}$ of ammonia.

Total mass of reactants and products is given as shown below:

$\begin{array}{c}\text{Total mass of reactants}={\text{Mass of N}}_{\text{2}}+{\text{Mass of H}}_{\text{2}}\\ =28.0\text{g}+6.0\text{g}\\ =\text{34}\text{.0}\text{g}\end{array}$

$\begin{array}{c}\text{Total mass of products}={\text{Mass of NH}}_{\text{3}}\\ =\text{34}\text{.0}\text{g}\end{array}$

Here, the mass of reactants is equal to mass of the products. Thus, law of conservation of mass is obeyed.

(b)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. It has to be interpreted in terms of particles, moles and mass and obeying of law of conservation of mass in it has to be shown.

$\text{HCl}\left(aq\right)+\text{KOH}\left(aq\right)\to \text{KCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Concept introduction:

According to law of conservation of mass, it is not possible to form the mass or to destroy the mass. The mass always remains conserved.

The number of moles is calculated as shown below:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

Answer to Problem 1PP

In the given equation shows that one molecule of aqueous hydrogen chloride reacts with one molecule of potassium hydroxide to form one molecule of potassium chloride and one molecule of water. In terms of number of moles, it is seen that one mole of aqueous hydrogen chloride reacts with one mole of potassium hydroxide to form one mole of potassium chloride and one mole of water. Here, $36.46\text{g}$ aqueous hydrogen chloride reacts with $56.106\text{g}$ of potassium hydroxide to form $18.0\text{g}$ of water and $74.55\text{g}$ of potassium chloride. The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below:

$\text{HCl}\left(aq\right)+\text{KOH}\left(aq\right)\to \text{KCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above equation shows that one molecule of aqueous hydrogen chloride reacts with one molecule of potassium hydroxide to form one molecule of potassium chloride and one molecule of water. In terms of number of moles, it is seen that one mole of aqueous hydrogen chloride reacts with one mole of potassium hydroxide to form one mole of potassium chloride and one mole of water.

The molar mass of hydrogen chloride is $\text{36}\text{.46}\text{g/mol}$. The mass of hydrogen chloride is calculated as shown below:

$\begin{array}{c}\text{1}\text{mol}=\frac{\text{Given mass}}{\text{36}\text{.46}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 36.46\frac{\text{g}}{\text{mol}}\\ =36.46\text{g}\end{array}$

The molar mass of potassium hydroxide is $\text{56}\text{.106}\text{g/mol}$. The mass of potassium hydroxide is calculated as shown below:

$\begin{array}{c}1\text{mol}=\frac{\text{Given mass}}{\text{56}\text{.106}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 56.106\frac{\text{g}}{\text{mol}}\\ =56.106\text{g}\end{array}$

The molar mass of potassium chloride is $\text{74}\text{.55}\text{g/mol}$. The mass of potassium chloride is calculated as shown below:

$\begin{array}{c}1\text{mol}=\frac{\text{Given mass}}{\text{74}\text{.55}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 74.55\frac{\text{g}}{\text{mol}}\\ =74.55\text{g}\end{array}$

The molar mass of water is $\text{18}\text{.0}\text{g/mol}$. The mass of water is calculated as shown below:

$\begin{array}{c}1\text{mol}=\frac{\text{Given mass}}{18.0\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 18.0\frac{\text{g}}{\text{mol}}\\ =18.0\text{g}\end{array}$

Thus, $36.46\text{g}$ aqueous hydrogen chloride reacts with $56.106\text{g}$ of potassium hydroxide to form $18.0\text{g}$ of water and $74.55\text{g}$ of potassium chloride.

Total mass of reactants and products is given as shown below:

$\begin{array}{c}\text{Total mass of reactants}=\text{Mass of HCl}+\text{Mass of KOH}\\ =36.46\text{g}+56.106\text{g}\\ =\text{92}\text{.566}\text{g}\end{array}$

$\begin{array}{c}\text{Total mass of products}=\text{Mass of KCl}+{\text{Mass of H}}_{\text{2}}\text{O}\\ =\text{74}\text{.55}\text{g+18}\text{.0}\text{g}\\ \text{=92}\text{.55}\text{g}\end{array}$

Here, the mass of reactants is equal to mass of the products. Thus, law of conservation of mass is obeyed.

(c)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. It has to be interpreted in terms of particles, moles and mass and obeying of law of conservation of mass in it has to be shown.

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

Concept introduction:

According to law of conservation of mass it is not possible to form the mass or to destroy the mass. The mass always remains conserved.

The number of moles is calculated as shown below:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

Answer to Problem 1PP

The given equation shows that two molecules of magnesium react with one molecule of oxygen to form two molecule of magnesium oxide. In terms of number of moles, it is seen that two moles of magnesium reacts with one mole of oxygen to form one mole of magnesium oxide. Here, $48.61\text{g}$ magnesium reacts with $32.0\text{g}$ of oxygen to form $80.6\text{g}$ of magnesium oxide. The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below:

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

The above equation shows that two molecules of magnesium react with one molecule of oxygen to form two molecule of magnesium oxide. In terms of number of moles it is seen that two moles of magnesium reacts with one mole of oxygen to form one mole of magnesium oxide.

The molar mass of magnesium is $\text{24}\text{.305}\text{g/mol}$. The mass of magnesium is calculated as shown below:

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{\text{24}\text{.305}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=2\text{mol}\times 24.305\frac{\text{g}}{\text{mol}}\\ =48.61\text{g}\end{array}$

The molar mass of oxygen gas is $\text{32}\text{.0}\text{g/mol}$. The mass of oxygen is calculated as shown below:

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{32.0\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 32.0\frac{\text{g}}{\text{mol}}\\ =32.0\text{g}\end{array}$

The molar mass of magnesium oxide is $\text{40}\text{.30}\text{g/mol}$. The mass of magnesium oxide is calculated as shown below:

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{\text{40}\text{.30}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=2\text{mol}\times 40.30\frac{\text{g}}{\text{mol}}\\ =80.6\text{g}\end{array}$

Thus, $48.61\text{g}$ magnesium reacts with $32.0\text{g}$ of oxygen to form $80.6\text{g}$ of magnesium oxide.

Total mass of reactants and products is given as shown below:

$\begin{array}{c}\text{Total mass of reactants}=\text{Mass of Mg}+{\text{Mass of O}}_{\text{2}}\\ =48.61\text{g}+32.0\text{g}\\ =\text{80}\text{.61}\text{g}\end{array}$

$\begin{array}{c}\text{Total mass of products}=\text{Mass of MgO}\\ =\text{80}\text{.6}\text{g}\end{array}$

Here, the mass of reactants is equal to mass of the products. Thus, law of conservation of mass is obeyed.