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Phosphoric acid $${H_3}P{O_4}$$ is neutralized by magnesium hydroxide, $$Mg{(OH)_2}$$. how many moles of water will be produced from the neutralization of 0.24 mole of $${H_3}P{O_4}$$?

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Hint: The neutralization is defined as the reaction in which the acid reacts with the bases or the alkali to produce water and salt as the product. In the reaction of neutralization the hydrogen ion of the acid gets combined with the hydroxyl ion of base to form water. The solution which is neutralised has the pH equal to 7 but it can differ on the basic and acidic strength of the reactants.

Complete step by step answer:
So to approach this question we should know the reaction that takes place between phosphoric acid and the magnesium hydroxide. Soon when phosphoric acid and magnesium hydroxide are reacted together the formation of magnesium orthophosphate and water takes place. The balanced equation for the reaction is:
$$3Mg{(OH)_2} + 2{H_3}P{O_4} \to M{g_3}{(P{O_4})_2} + 6{H_2}O$$
So from the reaction we can observe that,
2 moles of phosphoric acid produce = 6 moles of water so,
0.24 moles of phosphoric acid produce = $$\dfrac{6}{2} \times 0.24 = 0.72$$ moles of water

So 0.24 moles of phosphoric acid will produce 0.72 moles of water.

Note: The neutralization reaction is used in treating the wastewater so as to reduce the effluent created damage. It helps in determining the unknown concentration, the process of neutralization is used in chemical titration by analysing the bases and acids. The pH of the soil is also controlled by using the application of neutralization. The acidic nature of the soil can be controlled by adding calcium oxide to it.