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# How would you balance the following double replacement equation?$Al{(N{O_3})_3}\, + \,N{a_3}P{O_4}\, \to \,AlP{O_4}\, + \,NaN{O_3}$

Last updated date: 17th Jul 2024
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Hint:Here, as we have aluminum nitrate which reacts with sodium phosphate after the reaction we will leave with products as aluminum phosphate and sodium nitrate. 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 the same.

Complete step-by-step answer:We have the reaction taking place in between aluminum nitrate and sodium phosphate after the reaction we will leave with products as aluminum phosphate and sodium nitrate. Here, we know how these formulas get formed. In the case of phosphate it is charged as $PO_4^3$ and in case of nitrate it is shown as $NO_3^ -$ . Now lets see firstly how many atoms are present of each type on the left hand side and right hand side of reaction.
$Al{(N{O_3})_3}\, + \,N{a_3}P{O_4}\, \to \,AlP{O_4}\, + \,NaN{O_3} - - - - - - - (1)$
We have $(1Al,\,3Na,\,1P,\,3N\,and\,13O)$ on the left hand side of reaction and $(1Al,\,1Na,\,1P,\,1N\,and\,7O)$ on the right hand side of reaction. Let’s try to balance them first by multiplying the left hand side. This is because the left hand side has fewer atoms. We have sodium, nitrogen and oxygen unbalanced while all others are balanced. Now for balancing them we have to multiply sodium nitrate $NaN{O_3}$ by $3$ because after that the atoms of sodium and nitrogen become equal on both sides.
We get the reaction as, $Al{(N{O_3})_3}\, + \,N{a_3}P{O_4}\, \to \,AlP{O_4}\, + \,3NaN{O_3} - - - - - - - (2)\,Balanced\,chemical\,equation$
Now we have to count the number of atoms of each type on both side, $(1Al,\,3Na,\,1P,\,3N\,and\,13O)$ on left and $(1Al,\,3Na,\,1P,\,3N\,and\,13O)$. The equation is balanced now.

Note:As we see that the number of atoms on both sides becomes equal. There are various reactions which occur with the water of crystallization. So in those cases we have to eliminate the water of crystallization and then repeat the same process of balancing. At last we end with the same number of atoms at last.