Chapter 5, Problem 118E

Consider the following unbalanced equation:

${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_2\left(s\right)+{\text{ H}}_{\text{2}}{\text{SO}}_4\left(aq\right)\to {\text{CaSO}}_4\left(s\right)+{\text{ H}}_{\text{3}}{\text{PO}}_4\left(aq\right)$

What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H₂SO₄ by mass)?

Interpretation Introduction

Interpretation: The amount of calcium sulfate and phosphoric acid that can be produced from the stated reaction is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

To determine: The mass of Calcium sulfate and phosphoric acid that can be formed by the stated reaction.

Answer to Problem 118E

The mass of calcium sulfate and phosphoric acid formed is $\underset{\_}{1.3\times 10^{3}g}$ and $\underset{\_}{6.3\times 10^{2}g}$ respectively.

Explanation of Solution

To determine: The balanced form of the stated chemical equation.

According to the law of conservation of mass, mass can neither be created nor destroyed. The mass of reactants is equal to the mass of products formed. Therefore a chemical equation, having lesser number of moles of an element on either side of a reaction, is balanced using appropriate numerical coefficients to satisfy the law of conservation of mass.

The given reaction is,

${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{CaSO}}_4\left(s\right)+{\text{H}}_3{\text{PO}}_4\left(aq\right)$

Adding the coefficient $3$ to ${\text{CaSO}}_4$ balances the number of atoms of calcium present on either side of the reaction. The equation thus obtained is,

${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 3{\text{CaSO}}_4\left(s\right)+{\text{H}}_3{\text{PO}}_4\left(aq\right)$

Adding the coefficient $3$ to ${\text{H}}_2{\text{SO}}_4$ and the coefficient $2$ to ${\text{H}}_3{\text{PO}}_4$ balances the number of atoms of phosphorus, hydrogen and oxygen present on either side of the reaction. The balanced equation thus obtained is,

${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 3{\text{CaSO}}_4\left(s\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)$

Given

The balanced chemical reaction is,

${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 3{\text{CaSO}}_4\left(s\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)$

The given mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is $1.0\text{kg}\left(1000\text{g}\right)$.

The given mass of concentrated ${\text{H}}_{\text{2}}{\text{SO}}_4$ is $1.0\text{kg}\left(1000\text{g}\right)$.

The given amount of concentrated sulfuric acid is given to have $98\%$ sulfuric acid by mass.

Therefore, the mass of sulfuric acid $\begin{array}{l}=\left(\frac{98}{100}\times 1.0\right)\text{kg}\\ =0.98\text{kg}\end{array}$

The molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ $\begin{array}{l}=3\text{Ca}+2\text{P}+8\text{O}\\ =\left(\left(3\times 40\right)+\left(2\times 31\right)+\left(8\times 16\right)\right)\text{g}/\text{mol}\\ =310\text{g}/\text{mol}\end{array}$

The molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_4$ $\begin{array}{l}=2\text{H}+\text{S}+\text{4O}\\ =\left(\left(2\times 1\right)+32+\left(4\times 16\right)\right)\text{g}/\text{mol}\\ =98\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{the}\text{substance}}$

Substitute the value of the given mass and the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ and ${\text{H}}_{\text{2}}{\text{SO}}_4$ in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2=\frac{1000\text{kg}}{310\text{g}/\text{mol}}\\ =\text{3}\text{.2}\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_4=\frac{1000\text{kg}}{98\text{g}/\text{mol}}\\ =\text{10}\text{.2}\text{mol}\end{array}$

According to the stated reaction,

$1\text{mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ reacts with $3\text{mol}$ of ${\text{H}}_{\text{2}}{\text{SO}}_4$.

$\text{3}\text{.2}\text{mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ reacts with ${\text{H}}_{\text{2}}{\text{SO}}_4$ $\begin{array}{l}=\left(3\times \text{3}\text{.2}\right)\text{mol}\\ =9.6\text{mol}\end{array}$

The moles of ${\text{H}}_{\text{2}}{\text{SO}}_4$ given are $10.2\text{mol}$.

Therefore, some amount of ${\text{H}}_{\text{2}}{\text{SO}}_4$ is left unreacted. It is the excess reactant and ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is the limiting reactant.

According to the stated reaction,

$1\text{mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ produces $3\text{mol}$ of ${\text{CaSO}}_4$.

$\text{3}\text{.2}\text{mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ produce ${\text{CaSO}}_4$ $\begin{array}{l}=\left(3\times \text{3}\text{.2}\right)\text{mol}\\ =9.6\text{mol}\end{array}$

The molar mass of ${\text{CaSO}}_4$ $\begin{array}{l}=\text{Ca}+\text{S}+4\text{O}\\ =\left(40+132+\left(4\times 16\right)\right)\text{g}/\text{mol}\\ =136\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

Substitute the value of the number of moles of ${\text{CaSO}}_4$ and the molar mass of ${\text{CaSO}}_4$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}{\text{CaSO}}_4=\left(\text{9}\text{.6}\text{mol}\right)\times \left(136\text{g}/\text{mol}\right)\\ =\underset{\_}{1.3\times 10^{3}g}\end{array}$

Given

The balanced chemical reaction is,

${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+3{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 3{\text{CaSO}}_4\left(s\right)+2{\text{H}}_3{\text{PO}}_4\left(aq\right)$

According to the stated reaction,

$1\text{mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ produces $2\text{mol}$ of ${\text{H}}_3{\text{PO}}_4$.

$\text{3}\text{.2}\text{mol}$ of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ produce ${\text{H}}_3{\text{PO}}_4$ $\begin{array}{l}=\left(2\times \text{3}\text{.2}\right)\text{mol}\\ =6.4\text{mol}\end{array}$

The molar mass of ${\text{H}}_3{\text{PO}}_4$ $\begin{array}{l}=\text{3H}+\text{P}+4\text{O}\\ =\left(\left(3\times 1\right)+31+\left(4\times 16\right)\right)\text{g}/\text{mol}\\ =98\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

Substitute the value of the number of moles of ${\text{H}}_3{\text{PO}}_4$ and the molar mass of ${\text{H}}_3{\text{PO}}_4$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_3{\text{PO}}_4=\left(\text{6}\text{.4}\text{mol}\right)\times \left(98\text{g}/\text{mol}\right)\\ =\underset{\_}{6.3\times 10^{2}g}\end{array}$

Conclusion

The amount of calcium sulfate and phosphoric acid formed is $\underset{\_}{1.3\times 10^{3}g}$ and $\underset{\_}{6.3\times 10^{2}g}$ respectively.