Chapter 10, Problem 11PE

Interpretation Introduction

Interpretation:

The precipitated mass of calcium fluoride is to be calculated. The reactant which will be in excess is to be stated. The mass of the reactant that will be in excess is to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation, the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction. The ratio of moles is termed as mole ratio. In stoichiometry problems, the reactant that controls the amount of the product formed is known as the limiting reactant.

Answer to Problem 11PE

The precipitated mass of calcium fluoride is $\text{20}\text{.7}\text{g}$. The reactant which will be in excess is $\text{NaF}$. The mass of the reactant that will be in excess is $\text{17}\text{.}\text{3}\text{g}$.

Explanation of Solution

The balanced equation for the reaction is given below.

$\text{Ca}{\left({\text{NO}}_3\right)}_2\text{(aq)}+\text{2NaF}(\text{aq})\text{+}\to {\text{CaF}}_{\text{2}}\left(\text{s}\right)+2{\text{NaNO}}_{\text{3}}\text{(aq)}$

In the reaction, $2$ moles of $\text{NaF}$ react with $\text{1}$ mole of $\text{Ca}{\left({\text{NO}}_3\right)}_2$.

Therefore, mole to mole ratio is given below.

$\text{2}\text{mol}\text{NaF}=1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2$

Two conversion factors from the mole-to-mole ratio are given below.

$\frac{\text{2}\text{mol}\text{NaF}}{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}\text{and}\frac{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{2}\text{mol}\text{NaF}}$

The conversion factor to obtain moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ from $\text{NaF}$ is given below.

$\frac{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{2}\text{mol}\text{NaF}}$

The molar mass of sodium is $22.99\text{g}{\text{mol}}^{-1}$.

The molar mass of fluorine is $19.00\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{NaF}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 22.99\text{g}{\text{mol}}^{-1}+19.00\text{g}{\text{mol}}^{-1}\\ & =& 41.99\text{g}{\text{mol}}^{-1}\end{array}$

The formula to calculate the number of moles of $\text{NaF}$ is given below.

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

The mass of $\text{NaF}$ is $\text{39}\text{.5}\text{g}$.

Substitute the mass and molar mass of $\text{NaF}$ in equation (1).

$\begin{array}{rll}\text{Number}\text{of}\text{moles}\text{of}\text{NaF}& =& \frac{\text{39}\text{.5}\text{g}}{41.99\text{g}{\text{mol}}^{-1}}\\ & =& 0.941\text{mol}\end{array}$

The molar mass of calcium is $40.08\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 40.08\text{g}{\text{mol}}^{-1}+2\times \left(14.01\text{g}{\text{mol}}^{-1}+\left(3\times 16.00\text{g}{\text{mol}}^{-1}\right)\right)\\ & =& 40.08\text{g}{\text{mol}}^{-1}+2\times \left(14.01\text{g}{\text{mol}}^{-1}+48.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 40.08\text{g}{\text{mol}}^{-1}+124.02\text{g}{\text{mol}}^{-1}\\ & =& 164.1\text{g}{\text{mol}}^{-1}\end{array}$

The formula to calculate the number of moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is given below.

$\text{Number}\text{of}\text{moles}\text{of}\text{Ca}{\left({\text{NO}}_3\right)}_2=\frac{\text{Given}\text{mass}\text{of}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{Molar}\text{mass}\text{of}\text{Ca}{\left({\text{NO}}_3\right)}_2}$…(2)

The mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is $\text{43}\text{.5}\text{g}$.

Substitute the mass and molar mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ in equation (2).

$\begin{array}{rll}\text{Number}\text{of}\text{moles}\text{of}\text{Ca}{\left({\text{NO}}_3\right)}_2& =& \frac{\text{43}\text{.5}\text{g}}{164.1\text{g}{\text{mol}}^{-1}}\\ & =& 0.265\text{mol}\end{array}$

The number of moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ required to react with $0.941$ moles of $\text{NaF}$ is calculated below.

$\begin{array}{rll}\text{Moles}\text{of}\text{Ca}{\left({\text{NO}}_3\right)}_2& =& \frac{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{2}\text{mol}\text{NaF}}\times 0.941\text{mol}\text{NaF}\\ & =& 0.4705\text{mol}\end{array}$

The number of moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ present is $0.265\text{mol}$ which is less than the required number of moles. Therefore, $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is the limiting reactant.

Therefore, the number of moles of $\text{NaF}$ required to react with $0.265$ moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is calculated below.

$\begin{array}{rll}\text{Moles}\text{of}\text{NaF}& =& \frac{\text{2}\text{mol}\text{NaF}}{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}\times 0.265\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2\\ & =& 0.530\text{mol}\end{array}$

The moles of $\text{NaF}$ left are calculated below.

$\begin{array}{rll}\text{Unreacted}\text{moles}\text{of}\text{NaF}& =& 0.941\text{mol}-0.530\text{mol}\\ & =& \text{0}\text{.411}\text{mol}\end{array}$

The formula to calculate the mass of $\text{NaF}$ left is given below.

$\text{Mass}\text{of}\text{NaF}\text{left}=\text{Molar}\text{mass}\text{of}\text{NaF}\times \text{Number}\text{of}\text{moles}\text{of}\text{NaF}\text{left}$…(3)

Substitute molar mass and the mass of $\text{NaF}$ left in equation (3).

$\begin{array}{rll}\text{Mass}\text{of}\text{NaF}\text{left}& =& 41.99\text{g}{\text{mol}}^{-1}\times \text{0}\text{.411}\text{mol}\\ & =& 17\text{.}\text{3}\text{g}\end{array}$

Therefore, the mass of $\text{NaF}$ left is $\text{17}\text{.}\text{3}\text{g}$.

In the reaction, $1$ mole of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ produces $\text{1}$ mole of ${\text{CaF}}_{\text{2}}$.

Therefore mole to mole ratio is given below.

$1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2=1\text{mol}{\text{CaF}}_{\text{2}}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{1\text{mol}{\text{CaF}}_{\text{2}}}\text{and}\frac{1\text{mol}{\text{CaF}}_{\text{2}}}{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}$

The conversion factor to obtain moles of ${\text{CaF}}_{\text{2}}$ from $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is given below.

$\frac{1\text{mol}{\text{CaF}}_{\text{2}}}{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}$

Therefore, the number of moles of ${\text{CaF}}_{\text{2}}$ required to react with $0.265$ moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is calculated below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{CaF}}_{\text{2}}& =& \frac{1\text{mol}{\text{CaF}}_{\text{2}}}{1\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}\times 0.265\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2\\ & =& 0.265\text{mol}\end{array}$

The molar mass of calcium is $40.08\text{g}{\text{mol}}^{-1}$.

The molar mass of fluorine is $19.00\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{CaF}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 40.08\text{g}{\text{mol}}^{-1}+\left(2\times 19.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 78.08\text{g}{\text{mol}}^{-1}\end{array}$

The formula to calculate the mass of ${\text{CaF}}_{\text{2}}$ is given below.

$\text{Mass}\text{of}{\text{CaF}}_{\text{2}}=\text{Molar}\text{mass}\text{of}{\text{CaF}}_{\text{2}}\times \text{Number}\text{of}\text{moles}\text{of}{\text{CaF}}_{\text{2}}$…(4)

The number of moles of ${\text{CaF}}_{\text{2}}$ is $0.265\text{mol}$.

Substitute the moles and molar mass of ${\text{CaF}}_{\text{2}}$ in equation (4).

$\begin{array}{rll}\text{Mass}\text{of}{\text{CaF}}_{\text{2}}& =& 78.08\text{g}{\text{mol}}^{-1}\times 0.265\text{mol}\\ & =& \text{20}\text{.7}\text{g}\end{array}$

Therefore the mass of ${\text{CaF}}_{\text{2}}$ formed is $\text{20}\text{.7}\text{g}$.

After the reaction, the mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ will be completely used.

Therefore, the precipitated mass of calcium fluoride is $\text{20}\text{.7}\text{g}$. The reactant which will be in excess is $\text{NaF}$. The mass of the reactant that will be in excess is $\text{17}\text{.}\text{3}\text{g}$.

Conclusion

The precipitated mass of calcium fluoride is $\text{20}\text{.7}\text{g}$. The reactant which will be in excess is $\text{NaF}$. The mass of the reactant that will be in excess is $\text{17}\text{.}\text{3}\text{g}$.