Chapter 10, Problem 6PE

Interpretation Introduction

Interpretation:

The percentage yield when $\text{2}\text{.801}\text{g}$ of calcium phosphate is recovered from the reaction of a solution containing $\text{3}\text{.215}\text{g}$ of sodium phosphate with excess nitrate solution 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.

Answer to Problem 6PE

The percentage yield when $\text{2}\text{.801}\text{g}$ of calcium phosphate is recovered containing the reaction of a solution containing $\text{3}\text{.215}\text{g}$ of sodium phosphate with excess nitrate solution is $92.11\%$.

Explanation of Solution

The balanced equation for the reaction sodium phosphate with calcium nitrate solution is given below.

${\text{2Na}}_3{\text{PO}}_4\text{(aq)+3Ca}{\left({\text{NO}}_3\right)}_2\text{(aq)}\to 6{\text{NaNO}}_3\text{(aq)}+{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(\text{s}\right)$

Therefore, $2$ moles of ${\text{Na}}_3{\text{PO}}_4$ produce $\text{1}$ mole of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$.

Therefore mole to mole ratio is given below.

$\text{2}\text{mol}{\text{Na}}_3{\text{PO}}_4=1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$

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

$\frac{\text{2}\text{mol}{\text{Na}}_3{\text{PO}}_4}{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}\text{and}\frac{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{2}\text{mol}{\text{Na}}_3{\text{PO}}_4}$

The conversion factor to obtain moles of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ from ${\text{Na}}_3{\text{PO}}_4$ is given below.

$\frac{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{2}\text{mol}{\text{Na}}_3{\text{PO}}_4}$

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

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

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

Therefore, the molar mass of ${\text{Na}}_3{\text{PO}}_4$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(3\times 22.99\text{g}{\text{mol}}^{-1}\right)+30.97\text{g}{\text{mol}}^{-1}+\left(4\times \text{16}\text{.00}\text{g}{\text{mol}}^{-1}\right)\\ & =& \text{68}\text{.97}\text{g}{\text{mol}}^{-1}+30.97\text{g}{\text{mol}}^{-1}+64.00\text{g}{\text{mol}}^{-1}\\ & =& 163.94\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain moles of ${\text{Na}}_3{\text{PO}}_4$ from grams of ${\text{Na}}_3{\text{PO}}_4$ is given below.

$\frac{\text{1}\text{mol}{\text{Na}}_3{\text{PO}}_4}{\text{163}\text{.94}\text{g}{\text{Na}}_3{\text{PO}}_4}$

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

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

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

Therefore, the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(3\times 40.08\text{g}{\text{mol}}^{-1}\right)+2\times \left(30.97\text{g}{\text{mol}}^{-1}+\left(4\times \text{16}\text{.00}\text{g}{\text{mol}}^{-1}\right)\right)\\ & =& \text{120}\text{.24}\text{g}{\text{mol}}^{-1}+2\times \left(30.97\text{g}{\text{mol}}^{-1}+64.00\text{g}{\text{mol}}^{-1}\right)\\ & =& \text{120}\text{.24}\text{g}{\text{mol}}^{-1}+189.94\text{g}{\text{mol}}^{-1}\\ & =& 310.18\end{array}$

Therefore, the conversion factor to obtain grams of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ from moles of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is given below.

$\frac{\text{310}\text{.18}\text{g}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{1}\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}$

The formula to calculate the mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ from ${\text{Na}}_3{\text{PO}}_4$ is given below.

$\text{Mass}\text{of}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2=\left(\begin{array}{l}\text{Given}\text{mass}\text{of}\text{NO}\times \text{Conversion}\text{factor}\text{to}\text{obtain moles of }{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\\ \times \text{Conversion}\text{factor}\text{to obtain to}\text{obtain moles}\text{of}{\text{Na}}_3{\text{PO}}_4\\ \times \text{Conversion}\text{factor}\text{to obtain grams of}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\end{array}\right)$

The mass of ${\text{Na}}_3{\text{PO}}_4$ is $\text{3}\text{.125}\text{g}$.

Substitute the mass of ${\text{Na}}_3{\text{PO}}_4$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Mass}\text{of}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2& =& \left(\begin{array}{r}\text{3}\text{.215}\text{g}{\text{Na}}_3{\text{PO}}_4\times \frac{1\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{2}\text{mol}{\text{Na}}_3{\text{PO}}_4}\times \\ \frac{\text{1}\text{mol}{\text{Na}}_3{\text{PO}}_4}{\text{163}\text{.94}\text{g}{\text{Na}}_3{\text{PO}}_4}\times \frac{\text{310}\text{.18}\text{g}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}{\text{1}\text{mol}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2}\end{array}\right)\\ & =& \left(\begin{array}{c}\text{3}\text{.215}\text{g}{\text{Na}}_3{\text{PO}}_4\times 0.5\times \frac{\text{1}}{\text{163}\text{.94}\text{g}{\text{Na}}_3{\text{PO}}_4}\\ \times \text{310}\text{.18}\text{g}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\end{array}\right)\\ & =& 3.041\text{g}\end{array}$

The percentage yield is calculated by the formula given below.

$\text{Percentage}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Ideal}\text{yield}}\times 100$…(1)

The actual yield is $\text{2}\text{.801}\text{g}$.

The ideal yield is $3.041\text{g}$.

Substitute the value of actual yield and ideal yield in equation (1).

$\begin{array}{rll}\text{Percentage}\text{yield}& =& \frac{\text{2}\text{.801}\text{g}}{3.041\text{g}}\times 100\\ & =& 92.11\%\end{array}$

Therefore, the percentage yield when $\text{2}\text{.801}\text{g}$ of calcium phosphate is recovered form the reaction of a solution containing $\text{3}\text{.215}\text{g}$ of sodium phosphate with excess nitrate solution is $92.11\%$.

Conclusion

The percentage yield when $\text{2}\text{.801}\text{g}$ of calcium phosphate is recovered form the reaction of a solution containing $\text{3}\text{.215}\text{g}$ of sodium phosphate with excess nitrate solution is $92.11\%$.