Question

How many grams of phosphoric acid ${{H}_{3}}P{{O}_{4}}$ would be needed to neutralize 100 g of magnesium hydroxide $Mg{{(OH)}_{2}}$?

Answer 1

Hint: To answer this question firstly, we should know the reaction between ${{H}_{3}}P{{O}_{4}}$ and $Mg{{(OH)}_{2}}$. Being aware of the concept of limiting reagent will help us to solve this question within no time.

Complete Solution :
Firstly. Let's write the chemical equation of the reaction between ${{H}_{3}}P{{O}_{4}}$ and $Mg{{(OH)}_{2}}$:
 \[{{H}_{3}}P{{O}_{4}}+Mg{{(OH)}_{2}}\to Mg{{(P{{O}_{4}})}_{2}}+{{H}_{2}}O\]

- This problem can be solved with reference to the concept of the limiting reagent. Now, let's firstly know about the concept of limiting agent:
- In this reaction where more than one reactant is involved, one has first identified the limiting reagent that is the reactant which is completely consumed.
- To apply the concept of limiting reagent, the chemical equation should be balanced. So let's balance the chemical equation.
 \[2{{H}_{3}}P{{O}_{4}}+3Mg{{(OH)}_{2}}\to Mg{{(P{{O}_{4}})}_{2}}+3{{H}_{2}}O\]

- We should take the fraction of the number of moles divided by the stoichiometry of each reactant. The reactant which has the least number will be the limiting reagent.

- Let the number of moles of ${{H}_{3}}P{{O}_{4}}$ be ${{n}_{1}}$
${{H}_{3}}P{{O}_{4}}$ = $\dfrac{No.\text{ }of.\text{ }moles}{S.C}$ = $\dfrac{{{n}_{1}}}{2}$
Let the number of moles of $Mg{{(OH)}_{2}}$ be ${{n}_{2}}$
$Mg{{(OH)}_{2}}$= $\dfrac{No.\text{ }of.\text{ }moles}{S.C}$ = $\dfrac{{{n}_{2}}}{3}$
From this, it is evident that $Mg{{(OH)}_{2}}$ is the limiting reagent.

-Let me explain this let's assume that number of moles of ${{H}_{3}}P{{O}_{4}}$ and $Mg{{(OH)}_{2}}$ be 1. Always whenever the denominator is greater than the numerator then, the final value will be less. As the magnitude of the denominator increases, the final value decreases accordingly. This point is just an example taken to explain the limiting reagent.
The fraction of number of moles by stoichiometric number of limiting reagent is equal to the fraction of number of moles by stoichiometric of all the other reactants and products.

so,
\[\dfrac{{{n}_{2}}}{3} = \dfrac{{{n}_{1}}}{2}\]………………equation 1
Now, let find out number of moles of $Mg{{(OH)}_{2}}$= [${{n}_{2}}$]
Number of moles = $\dfrac{~Weight\text{ }in\text{ }g~~~~}{Molecular\text{ }weight}$
Weight of $Mg{{(OH)}_{2}}$ is 100 g (Given)
Molecular weight of$Mg{{(OH)}_{2}}$is 58.3
\[{{n}_{2}} = \dfrac{100}{58.3}\]
${{n}_{2}}$ = 1.7152
Substituting ${{n}_{2}}$ value in equation 1
\[\dfrac{1.7152}{3} = \dfrac{{{n}_{1}}}{2}\]
${{n}_{1}}$ = $\dfrac{1.7152\times 2}{3}$
${{n}_{1}}$ = 1.1434

The number of moles of ${{H}_{3}}P{{O}_{4}}$ is 1.1434
We know that,
Number of moles = $\dfrac{~Weight\text{ }in\text{ }g~~~~}{Molecular\text{ }weight}$………….equation 2
Molecular weight of ${{H}_{3}}P{{O}_{4}}$= 98 g
Substituting the number of moles of ${{H}_{3}}P{{O}_{4}}$ and the molecular weight in order to find the weight of ${{H}_{3}}P{{O}_{4}}$.

- Let ' W' be the weight of ${{H}_{3}}P{{O}_{4}}$
1.1434 = $\dfrac{W}{98}$
W = 1.1434 $\times $ 98
W = 112.05
The weight of ${{H}_{3}}P{{O}_{4}}$ is 112.05 g

Note: Understanding the concept of limiting reagent helps us to solve within no time as the calculations will be less and it is easy, but make sure that the chemical equation is balanced while applying this method. This can be worked out by applying POAC consequently and finding the value but it consumes lots of time and there will be lots of calculation.