How do you balance Mg(NO3)2+K3PO4→Mg3(PO4)2+KNO3?

2 Answers
anor277
Mar 7, 2017

3Mg(NO3)2+2K3PO4→Mg3(PO4)2+6KNO3

Explanation:

Is the given reaction stoichiometrically balanced? For every reactant particle, is there a corresponding product particle. If there is not, then the reaction cannot be accepted as a representation of chemical (i.e. physical) reality.

Nikka C.
Mar 10, 2017

3Mg(NO3)2 + 2K3PO4 → Mg3(PO4)2 + )6KNO3

Explanation:

Let's assume that the chemical reaction will progress as written.

Balancing ionic equations can be quite easy AS LONG AS YOU MEMORIZE YOUR CATIONS AND ANIONS. There are a number of ways to do this, so you have to find a way that actually works for you. For now, let's do a tally sheet of all ions involved.

Mg(NO3)2 + K3PO4 → Mg3(PO4)2 + KNO3

Left side:
Mg2+ = 1
NO−13 = 2
K1+ = 3
PO−34 = 1

Right side:

Mg2+ = 3
NO−13 = 1
K1+ = 1
PO−34 = 2

Let's balance the most complicated ion first: the PO3−4.

Left side:
Mg2+ = 1
NO−13 = 2
K1+ = 3
PO−34 = 1 x 2 = 2

Right side:

Mg2+ = 3
NO−13 = 1
K1+ = 1
PO−34 = 2

Since the PO3−4 ion is bonded to the K1+ ion, you need to multiply it by 2 as well.

Left side:
Mg2+ = 1
NO−13 = 2
K1+ = 3 x 2 = 6
PO−34 = 1 x 2 = 2

Right side:

Mg2+ = 3
NO−13 = 1
K1+ = 1
PO−34 = 2

Mg(NO3)2 + 2K3PO4 → Mg3(PO4)2 + KNO3

Now notice that this move have created an imbalance in your K1+ ion with 6 ions on the left as opposed to only one ion on the right. To compensate, we have:

Left side:
Mg2+ = 1
NO−13 = 2
K1+ = 3 x 2 = 6
PO−34 = 1 x 2 = 2

Right side:

Mg2+ = 3
NO−13 = 1
K1+ = 1 x 6 = 6
PO−34 = 2

Again, since the K1+ ion is bonded to the NO−13, we need to also multiply it by 6.

Left side:
Mg2+ = 1
NO−13 = 2
K1+ = 3 x 2 = 6
PO−34 = 1 x 2 = 2

Right side:

Mg2+ = 3
NO−13 = 1 x 6 = 6
K1+ = 1 x 6 = 6
PO−34 = 2

Mg(NO3)2 + 2K3PO4 → Mg3(PO4)2 + 6KNO3

Now your NO−13 ion is unbalanced with only 2 ions on the left as opposed to the 6 on the right. Thus,

Left side:
Mg2+ = 1
NO−13 = 2 x 3 = 6
K1+ = 3 x 2 = 6
PO−34 = 1 x 2 = 2

Right side:

Mg2+ = 3
NO−13 = 1 x 6 = 6
K1+ = 1 x 6 = 6
PO−34 = 2

Same as above the NO−13 ion is bonded with your Mg2+. Therefore,

Left side:
Mg2+ = 1 x 3 ==3
NO−13 = 2 x 3 = 6
K1+ = 3 x 2 = 6
PO−34 = 1 x 2 = 2

Right side:

Mg2+ = 3
NO−13 = 1 x 6 = 6
K1+ = 1 x 6 = 6
PO−34 = 2

Final Answer:

3Mg(NO3)2 + 2K3PO4 → Mg3(PO4)2 + color(blue)6KNO_3 (balanced)

Related questions
See all questions in Balancing Chemical Equations
Impact of this question
2376 views around the world
You can reuse this answer
Creative Commons License