# What is the equation of reaction between magnesium nitrate and ammonium hydrogen phosphate??

Feb 17, 2018

$\text{Mg"("NO"_3)_2"(aq)" + ("NH"_4)_2"HPO"_4"(aq)" + "NH"_3"(aq)" +"6H"_2"O(l)}$
→ "Mg"("NH"_4)"PO"_4⋅6"H"_2"O(s)" +"2NH"_4"NO"_3(aq)"

#### Explanation:

This reaction is a precipitation reaction — where the cation and the anion of one compound pair up with the cation and anion of another aqueous compound, producing a solid precipitate.

First, we need to know our reactants.

• Magnesium nitrate would be $M {g}^{2 +}$ and $N {O}_{3}^{-}$. Together, they must be $M g {\left(N {O}_{3}\right)}_{2}$ to balance charge.
Magnesium nitrate is soluble in water because it contains the nitrate anion.
• Ammonium hydrogen phosphate is $N {H}_{4}^{+}$ and $H P {O}_{4}^{2 -}$. Together, they must be ${\left(N {H}_{4}\right)}_{2} H P {O}_{4}$.
Ammonium hydrogen phosphate is soluble in water because it contains the ammonium cation.

However, the reaction also uses aqueous ammonia, which is a base. It reacts with the acidic hydrogen phosphate ion.

$\text{NH"_3 + "HPO"_4^"2-" ⇌ "NH"_4^"+" + "PO"_4^"3-}$

To find our products, we just need to "change the partners" of our cation and anion.

• $M {g}^{2 +}$ and $\text{NH"_4^"+}$ will go with $\text{PO"_4^"3-}$, making ${\text{Mg"("NH"_4)"PO}}_{4}$. The product is the hexahydrate, $\text{Mg"("NH"_4)"PO"_4⋅6"H"_2"O}$
It is insoluble in water, so that means it will form a precipitate.

• $N {H}_{4}^{+}$ will go with $N {O}_{3}^{-}$, making $N {H}_{4} N {O}_{3}$.
Ammonium nitrate is soluble in water.

After figuring out the reactants and the products, here's our net ionic equation:

$\text{Mg"^"2+" + "NH"_3 + "HPO"_4^"-" + "H"_2"O" → "Mg"("NH"_4)"PO"_4⋅6"H"_2"O}$

This reaction is a precipitation reaction, so we should add states of matter. And, since this reaction is taking place in water, we can add states of matter by labelling compounds which are soluble in water with $\left(a q\right)$, indicating that they're aqueous.

The precipitate, which is insoluble in water, would be marked with $\left(s\right)$, indicating that it is a solid.

$\text{Mg"^"2+""(aq)" + "NH"_3"(aq)" + "HPO"_4^"-""(aq)" + "6H"_2"O(l)" → "Mg"("NH"_4)"PO"_4⋅6"H"_2"O(s)}$

When we add the spectator ions, we get

$\text{Mg"("NO"_3)_2"(aq)" + ("NH"_4)_2"HPO"_4"(aq)" + "NH"_3"(aq)" +"6H"_2"O(l)}$
→ "Mg"("NH"_4)"PO"_4⋅6"H"_2"O(s)" +"2NH"_4"NO"_3"(aq)"