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# How would you balance the following equation? ${H_3}P{O_4}\, + \,Mg{(OH)_2}\, \to \,M{g_3}{(P{O_4})_2}\, + \,{H_2}O$

Last updated date: 18th Jun 2024
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Hint:While balancing a chemical equation we have to keep in mind that the number of atoms on each side should be equal. We can multiply the stoichiometric coefficient of each reactant and product and make each type of atom same. Here, atoms are hydrogen oxygen magnesium phosphorus. We have found that equation in which each atom is balanced at last.

Complete step-by-step answer:We have the reaction taking place in between magnesium hydroxide and phosphoric acid. Magnesium is an element of the group $2$ so if it reacts with water it makes hydroxide of formula $Mg{(OH)_2}$ while phosphoric acid has an oxidation number of phosphorus as $+ 5$ . They reacts and forms magnesium phosphate and release water in the solution.
${H_3}P{O_4}\, + \,Mg{(OH)_2}\, \to \,M{g_3}{(P{O_4})_2}\, + \,{H_2}O\, - - - - - - - - - - - - - \,(1)$
Count the number of atoms left hand side and right hand side. We have atoms on left hand side as $(\,1P,\,5H,\,6O\,and\,1Mg)$ here just symbols are used to define the numbers of atoms, now lets’ count on right hand side and we have $(\,2P,\,2H,\,9O\,and\,3Mg)$. Now here we have four species, let’s try to balance them first by multiplying the left hand side. This is because the left hand side has fewer atoms. Let’s start by multiplying phosphoric acid by number $2$ and magnesium hydroxide by number $3$ we get the equation as
$2{H_3}P{O_4}\, + \,3Mg{(OH)_2}\, \to \,M{g_3}{(P{O_4})_2}\, + \,{H_2}O\, - - - - - - - - - - - - - (2)$ .
To confirm the change in number of atoms let’s again count the atoms left hand side have $(2P,\,12H,\,14O\,and\,3Mg)$ and right hand side have $(\,2P,\,2H,\,9O\,and\,3Mg)$. Now as the number on the right hand side decreases we have to multiply the right hand side. We can multiply the water molecule by number $6$ . The equation becomes as-
$2{H_3}P{O_4}\, + \,3Mg{(OH)_2}\, \to \,M{g_3}{(P{O_4})_2} + \,6{H_2}O\, - - - - - - - - - - - - (3)$
If you count the number of atoms on both sides it becomes equal. Both sides we have two phosphorus, hydrogens are twelve, oxygen are fourteen in number and at last we have three magnesium.
Thus the balanced chemical equation is $2{H_3}P{O_4}\, + \,3Mg{(OH)_2}\, \to \,M{g_3}{(P{O_4})_2} + \,6{H_2}O\, - - - - (Final\,answer)$

Note:This is a neutralization reaction of an acid with base, after combination they form salt and water. The reaction is exothermic in nature; it means that energy is released in a larger amount. The balancing of chemical equations is very important because by this process we get to know the moles of reactant used and products formed.