Question

One equivalent of calcium phosphate $C{a_3}{\left( {P{O_4}} \right)_2}$ is dissolved in $HCl$ as,
$C{a_3}{\left( {P{O_4}} \right)_2}{\text{ + 6HCl }} \to {\text{ 3CaC}}{{\text{l}}_2}{\text{ + 2}}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$
The moles of $CaC{l_2}$ formed are $\dfrac{1}{x}$ , then the value of $x$ is:

Answer 1

For finding the value of $x$ we will first find the equivalent weight of calcium phosphate $C{a_3}{\left( {P{O_4}} \right)_2}$ , then by using the concepts of stoichiometry we will find the number of moles of Calcium Chloride $CaC{l_2}$ formed during the reaction. Thus we can find the value of $x$ which is the reciprocal of the number of moles of $CaC{l_2}$ formed.
Formula Used:
Equivalent weight $= {\text{ }}\dfrac{{{\text{ Molecular weight}}}}{{n - factor}}$

Complete answer:
The dissociation of calcium phosphate can be represented as:
$C{a_3}{\left( {P{O_4}} \right)_2}{\text{ }} \rightleftharpoons {\text{ 3C}}{{\text{a}}^{2 + }}{\text{ + 2P}}{{\text{O}}_4}^{3 - }$
Thus the oxidation state of metal atoms is two. But as we see, three moles of calcium atoms are involved in calcium phosphate compounds. Therefore the total valency of calcium is:
${\text{ 3 }} \times {\text{ 2 = 6}}$
$n -$factor can be also represented by the valency of the metal atom of the compound. Now we will find the equivalent weight of calcium phosphate as,
Equivalent weight $= {\text{ }}\dfrac{{{\text{ Molecular weight}}}}{{n - factor}}$
Equivalent weight of calcium phosphate $= {\text{ }}\dfrac{{{\text{ Molecular weight}}}}{6}$
It can be also written as,
One equivalent of $C{a_3}{\left( {P{O_4}} \right)_2}{\text{ = }}\dfrac{{{\text{number of moles}}}}{6}$
From the concept of stoichiometry we can observe from the above reaction that one mole of calcium phosphate produces three moles of calcium chloride and two moles of phosphoric acid. This can be also represented as:
$1{\text{ mole C}}{{\text{a}}_3}{\left( {P{O_4}} \right)_2}{\text{ = 3 mole CaC}}{{\text{l}}_2}{\text{ = 2 moles }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$
On dividing each term by taking L.C.M of $(1{\text{ , 2 , 3}})$ we get the result as,
$\dfrac{1}{6}{\text{ mole C}}{{\text{a}}_3}{\left( {P{O_4}} \right)_2}{\text{ = }}\dfrac{{\text{3}}}{6}{\text{ mole CaC}}{{\text{l}}_2}{\text{ = }}\dfrac{{\text{2}}}{6}{\text{ moles }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$
It can be simplified as,
$\dfrac{1}{6}{\text{ mole C}}{{\text{a}}_3}{\left( {P{O_4}} \right)_2}{\text{ = }}\dfrac{1}{2}{\text{ mole CaC}}{{\text{l}}_2}{\text{ = }}\dfrac{1}{3}{\text{ moles }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$
According to question moles of $CaC{l_2}$ formed are $\dfrac{1}{x}$ and on comparing we get the value of $x$ as 2.

Note:
We have solved this question by using the unitary method. We have calculated the equivalent of each compound in the reaction and then found the moles of calcium chloride. This method is valid for all reactions. Also we can determine the moles of phosphoric acid too.