Chapter 11.1, Problem 2PP

(a)

Interpretation Introduction

Interpretation:

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

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

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 2PP

In the given equation shows that two molecules of sodium reacts with two molecules of water to form two molecules of sodium hydroxide and one molecule of hydrogen gas. In terms of number of moles it is seen that two moles of sodium reacts with two moles of water gas to form two moles of sodium hydroxide and one mole of hydrogen gas. Here, $46.0\text{g}$ sodium reacts with $36.0\text{g}$ of water to form $80.0\text{g}$ of sodium hydroxide and $2.0\text{g}$ of hydrogen.

The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below.

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

The coefficient of two is multiplied with sodium, water and sodium hydroxide to balance the chemical equation.

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

The above equation shows that two molecules of sodium reacts with two molecules of water to form two molecules of sodium hydroxide and one molecule of hydrogen gas. In terms of number of moles it is seen that two moles of sodium reacts with two moles of water gas to form two moles of sodium hydroxide and one mole of hydrogen gas.

The molar mass of sodium is $\text{23}\text{.0}\text{g/mol}$. The mass of sodium is calculated as shown below.

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{\text{23}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=2\text{mol}\times 23.\text{0}\frac{\text{g}}{\text{mol}}\\ =46.0\text{g}\end{array}$

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

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{\text{18}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=2\text{mol}\times 18.\text{0}\frac{\text{g}}{\text{mol}}\\ =36.0\text{g}\end{array}$

The molar mass of sodium hydroxide is $\text{40}\text{.0}\text{g/mol}$. The mass of sodium hydroxide is calculated as shown below.

$\begin{array}{c}2\text{mol}=\frac{\text{Given mass}}{\text{40}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=2\text{mol}\times 40.\text{0}\frac{\text{g}}{\text{mol}}\\ =80.0\text{g}\end{array}$

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

$\begin{array}{c}1\text{mol}=\frac{\text{Given mass}}{\text{2}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 2.\text{0}\frac{\text{g}}{\text{mol}}\\ =2.0\text{g}\end{array}$

Thus, $46.0\text{g}$ sodium reacts with $36.0\text{g}$ of water to form $80.0\text{g}$ of sodium hydroxide and $2.0\text{g}$ of hydrogen.

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

$\begin{array}{c}\text{Total mass of reactants}=\text{Mass of Na}+{\text{Mass of H}}_{\text{2}}\text{O}\\ =46.0\text{g}+36.0\text{g}\\ =\text{82}\text{.0}\text{g}\end{array}$

$\begin{array}{c}\text{Total mass of products}=\text{Mass of NaOH}+{\text{Mass of H}}_{\text{2}}\\ =\text{80}\text{.0}\text{g}+\text{2}\text{.0}\text{g}\\ =\text{82}\text{.0}\text{g}\end{array}$

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

$\text{Zn}\left(s\right)+{\text{HNO}}_3\left(aq\right)\to \text{Zn}{\left({\text{NO}}_3\right)}_2\left(aq\right)+{\text{N}}_{\text{2}}\text{O}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 2PP

In the given equation shows that four molecules of zinc reacts with ten molecules of nitric acid to form four molecules of zinc nitrate, one molecule of nitrous oxide and five molecules of water. In terms of number of moles it is seen that four moles of zinc reacts with ten moles of nitric acid to form four moles of zinc nitrate, one mole of nitrous oxide and five moles of water. Here, $261.52\text{g}$ zinc reacts with $630.1\text{g}$ of nitric acid to form $757.44\text{g}$ of zinc nitrate, $44.01\text{g}$ of nitrous oxide and $90.0\text{g}$ of water.

The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below.

$\text{Zn}\left(s\right)+{\text{HNO}}_3\left(aq\right)\to \text{Zn}{\left({\text{NO}}_3\right)}_2\left(aq\right)+{\text{N}}_{\text{2}}\text{O}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The coefficient of four is multiplied with zinc, ten with nitric acid, four with zinc nitrate and five with water to balance the given chemical equation.

$\text{4Zn}\left(s\right)+10{\text{HNO}}_3\left(aq\right)\to 4\text{Zn}{\left({\text{NO}}_3\right)}_2\left(aq\right)+{\text{N}}_{\text{2}}\text{O}\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above equation shows that four molecules of zinc reacts with ten molecules of nitric acid to form four molecules of zinc nitrate, one molecule of nitrous oxide and five molecules of water. In terms of number of moles it is seen that four moles of zinc reacts with ten moles of nitric acid to form four moles of zinc nitrate, one mole of nitrous oxide and five moles of water.

The molar mass of zinc is $\text{65}\text{.38}\text{g/mol}$. The mass of zinc is calculated as shown below.

$\begin{array}{c}4\text{mol}=\frac{\text{Given mass}}{65.38\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=4\text{mol}\times 65.38\frac{\text{g}}{\text{mol}}\\ =261.52\text{g}\end{array}$

The molar mass of nitric acid is $\text{63}\text{.01}\text{g/mol}$. The mass of nitric acid is calculated as shown below.

$\begin{array}{c}10\text{mol}=\frac{\text{Given mass}}{63.01\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=10\text{mol}\times 63.01\frac{\text{g}}{\text{mol}}\\ =630.1\text{g}\end{array}$

The molar mass of zinc nitrate is $\text{189}\text{.36}\text{g/mol}$. The mass of zinc nitrate is calculated as shown below.

$\begin{array}{c}4\text{mol}=\frac{\text{Given mass}}{189.36\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=4\text{mol}\times 189.36\frac{\text{g}}{\text{mol}}\\ =757.44\text{g}\end{array}$

The molar mass of nitrous oxide is $\text{44}\text{.01}\text{g/mol}$. The mass of nitrous oxide is calculated as shown below.

$\begin{array}{c}1\text{mol}=\frac{\text{Given mass}}{44.01\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1\text{mol}\times 44.01\frac{\text{g}}{\text{mol}}\\ =44.01\text{g}\end{array}$

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

$\begin{array}{c}5\text{mol}=\frac{\text{Given mass}}{18.0\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=5\text{mol}\times 18.0\frac{\text{g}}{\text{mol}}\\ =90.0\text{g}\end{array}$

Thus, $261.52\text{g}$ zinc reacts with $630.1\text{g}$ of nitric acid to form $757.44\text{g}$ of zinc nitrate, $44.01\text{g}$ of nitrous oxide and $90.0\text{g}$ of water.

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

$\begin{array}{c}\text{Total mass of reactants}=\text{Mass of Zn}+{\text{Mass of HNO}}_{\text{3}}\\ =261.52\text{g}+630.1\text{g}\\ =\text{891}\text{.62}\text{g}\end{array}$

$\begin{array}{c}\text{Total mass of products}=\text{Mass of Zn}{\left({\text{NO}}_3\right)}_2+{\text{Mass of N}}_{\text{2}}\text{O}+{\text{Mass of H}}_{\text{2}}\text{O}\\ =\text{757}\text{.44}\text{g}+\text{44}\text{.01}\text{g}+90.0\text{g}\\ =\text{891}\text{.45}\text{g}\end{array}$

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