Chapter 9, Problem 30E

Interpretation Introduction

Interpretation:

The mass of calcium hydroxide that reacts to give $\text{2}\text{.39}\text{g}$ of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is to be calculated.

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, a known amount of product or reactant is given and the other product or reactant has to be calculated.

Answer to Problem 30E

The mass of calcium hydroxide that reacts to give $\text{2}\text{.39}\text{g}$ of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $1.71\text{g}$.

Explanation of Solution

The reaction is given below.

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+\text{Ca}{\left(\text{OH}\right)}_{\text{2}}(aq)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\text{NaOH}(aq)$

The balanced equation for the above reactsion is given below.

${\text{2Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+3\text{Ca}{\left(\text{OH}\right)}_{\text{2}}(aq)\to {\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+6\text{NaOH}(aq)$

In the reactsion, $3$ moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ produce $1$ mole of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

Therefore, the mole ratio is given below.

$\frac{3\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\text{and}\frac{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}{3\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}$

The mole ratio to obtain moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ from moles of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is given below.

$\frac{3\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}$

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

The molar mass of phosphorus is $30.97\text{g}{\text{mol}}^{-1}$.

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

Therefore, the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(3\times 40.08\text{g}{\text{mol}}^{-1}\right)+2\times \left(30.97\text{g}{\text{mol}}^{-1}+\left(4\times 16.00\text{g}{\text{mol}}^{-1}\right)\right)\\ & =& 120.24\text{g}{\text{mol}}^{-1}+2\times \left(30.97\text{g}{\text{mol}}^{-1}+64.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 120.24\text{g}{\text{mol}}^{-1}+189.94\text{g}{\text{mol}}^{-1}\\ & =& 310.18\text{g}{\text{mol}}^{-1}\end{array}$

The conversion factor to calculate moles of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ from grams of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is shown below.

$\frac{\text{1}\text{mol}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}{310.18\text{g}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}$

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

The molar mass of hydrogen is $1.01\text{g}{\text{mol}}^{-1}$.

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

Therefore, the molar mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(40.08\text{g}{\text{mol}}^{-1}\right)+\left(2\times \left(1.01\text{g}{\text{mol}}^{-1}+16.00\text{g}{\text{mol}}^{-1}\right)\right)\\ & =& 40.08\text{g}{\text{mol}}^{-1}+2\times \left(17.01\text{g}{\text{mol}}^{-1}\right)\\ & =& 40.08\text{g}{\text{mol}}^{-1}+34.02\text{g}{\text{mol}}^{-1}\\ & =& 74.1\text{g}{\text{mol}}^{-1}\end{array}$

The conversion factor to calculate grams of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ from moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is shown below.

$\frac{74.1\text{g}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{1}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}$

The formula to calculate the mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ from the mass of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is given below.

$\text{Mass}\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}=\left(\begin{array}{l}\text{Given}\text{mass}\text{of}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\\ \times \text{Conversion}\text{factor}\text{to}\text{obtain}\text{moles}\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\\ \times \text{Conversion}\text{factor}\text{to}\text{obtain}\text{moles}\text{of}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\\ \times \text{Conversion}\text{factor}\text{to}\text{obtain}\text{grams}\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\end{array}\right)$  …(1)

The mass of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{2}\text{.39}\text{g}$.

Substitute the value of mass of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$, mole ratio and conversion factors in equation (1).

$\begin{array}{rll}\text{Mass}\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}& =& \left(\begin{array}{l}\text{2}\text{.39}\text{g}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\times \frac{3\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\times \frac{\text{1}\text{mol}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}{310.18\text{g}{\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}}\\ \times \frac{74.1\text{g}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{1}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}\end{array}\right)\\ & =& 1.71\text{g}\end{array}$

Therefore, the mass of calcium hydroxide that reacts to give $\text{2}\text{.39}\text{g}$ of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is

$1.71\text{g}$.

Conclusion

The mass of calcium hydroxide that reacts to give $\text{2}\text{.39}\text{g}$ of ${\text{Ca}}_3{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $1.71\text{g}$.