# What is the balanced equation for this reaction: ?K_3PO_4 + ?Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3?

Dec 10, 2016

$2 {K}_{3} P {O}_{4} + 3 C a {\left(N {O}_{3}\right)}_{2} \to C {a}_{3} {\left(P {O}_{4}\right)}_{2} + 6 K N {O}_{3}$

#### Explanation:

So there's not really a whole lot of concepts to explain here. However, one handy thing to keep in mind here is to treat polyatomic ions as individual atoms. This will save you a load of time, and makes visualizing the scenario much less trickier.

Since there's really no one correct mathematical way to do this problem, I'm just going to briefly walk you through my thought process as I did it:

My first idea was to try and get the same number of calciums on both sides. To do this, I added a 3 in front of $C a {\left(N {O}_{3}\right)}_{2}$ This took care of the calciums, but left me with $6 \left(N {O}_{3}\right)$s as opposed to 1 on the other:

?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + ?KNO_3?

To even this out, I added a 6 in front of the $K N {O}_{3}$. Now, however, I had 6 Potassiums on one side as opposed to 3 on the other:

?K_3PO_4 + 3Ca(NO_3)_2 -> ?Ca_3(PO_4)_2 + 6KNO_3?

To even this out, I added a 2 in front of the ${K}_{3} P {O}_{4}$, which took care of the Potassiums, and also evened out the number of $\left(P {O}_{4}\right)$s on both sides. This left me with my final answer:

$2 {K}_{3} P {O}_{4} + 3 C a {\left(N {O}_{3}\right)}_{2} \to C {a}_{3} {\left(P {O}_{4}\right)}_{2} + 6 K N {O}_{3}$

As you can see, there's no one right way to solve these problems. Rather, it's just about fixing one thing, and seeing where that leads you.

Hope that helped :)