How do you balance #Mg(NO_3)_2 + K_3PO_4 - Mg_3(PO_4)_2 + KNO_3#?

Question

How do you balance #Mg(NO_3)_2 + K_3PO_4 -> Mg_3(PO_4)_2 + KNO_3#?

Mateo Aguilar · Accepted Answer

#3Mg(NO_3)_2 + 2K_3PO_4 rarr Mg_3(PO_4)_2 + 6KNO_3#

Is the reaction that has been described stoichiometrically balanced? Is there a corresponding product particle for each reactant particle? If not, the reaction cannot be accepted as a representation of chemical (or physical) reality.

Peyton Adams · Answer

#3Mg(NO_3)_2# + #2K_3PO_4# #rarr# #Mg_3(PO_4)_2# + #)6KNO_3#

Assuming the chemical reaction proceeds as described, let's proceed.

Ionic equation balancing can be fairly simple AS LONG AS YOU MEMORIZE YOUR CATIONS AND ANIONS. There are several strategies for this, so you'll need to figure out what works best for you. For the purposes of this discussion, let's create a tally sheet of all the ions that are involved.

#Mg(NO_3)_2# + #K_3PO_4# #rarr# #Mg_3(PO_4)_2# + #KNO_3#

Left side: #Mg^(2+)# = 1 #NO_3^(-1)# = 2 #K^(1+)# = 3 #PO_4^(-3)# = 1

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 #K^(1+)# = 1 #PO_4^(-3)# = 2

Let's balance the most complicated ion first: the #PO_4^(3-)#.

Left side: #Mg^(2+)# = 1 #NO_3^(-1)# = 2 #K^(1+)# = 3 #PO_4^(-3)# = 1 x #color(red)2# = 2

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 #K^(1+)# = 1 #PO_4^(-3)# = 2

Since the #PO_4^(3-)# ion is bonded to the #K^(1+)# ion, you need to multiply it by 2 as well.

Left side: #Mg^(2+)# = 1 #NO_3^(-1)# = 2 #K^(1+)# = 3 x #color(red)2# = 6 #PO_4^(-3)# = 1 x #color(red)2# = 2

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 #K^(1+)# = 1 #PO_4^(-3)# = 2

#Mg(NO_3)_2# + #color(red)2K_3PO_4# #rarr# #Mg_3(PO_4)_2# + #KNO_3#

Now notice that this move have created an imbalance in your #K^(1+)# ion with 6 ions on the left as opposed to only one ion on the right. To compensate, we have:

Left side: #Mg^(2+)# = 1 #NO_3^(-1)# = 2 #K^(1+)# = 3 x 2 = 6 #PO_4^(-3)# = 1 x 2 = 2

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 #K^(1+)# = 1 x #color(blue)6# = 6 #PO_4^(-3)# = 2

Again, since the #K^(1+)# ion is bonded to the #NO_3^(-1)#, we need to also multiply it by 6.

Left side: #Mg^(2+)# = 1 #NO_3^(-1)# = 2 #K^(1+)# = 3 x 2 = 6 #PO_4^(-3)# = 1 x 2 = 2

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 x #color(blue)6# = 6 #K^(1+)# = 1 x #color(blue)6# = 6 #PO_4^(-3)# = 2

#Mg(NO_3)_2# + #color(red)2K_3PO_4# #rarr# #Mg_3(PO_4)_2# + #color(blue)6KNO_3#

Now your #NO_3^(-1)# ion is unbalanced with only 2 ions on the left as opposed to the 6 on the right. Thus,

Left side: #Mg^(2+)# = 1 #NO_3^(-1)# = 2 x #color(green)3# = 6 #K^(1+)# = 3 x 2 = 6 #PO_4^(-3)# = 1 x 2 = 2

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 x 6 = 6 #K^(1+)# = 1 x 6 = 6 #PO_4^(-3)# = 2

Same as above the #NO_3^(-1)# ion is bonded with your #Mg^(2+)#. Therefore,

Left side: #Mg^(2+)# = 1 x #color(green)3# ==3 #NO_3^(-1)# = 2 x #color(green)3# = 6 #K^(1+)# = 3 x 2 = 6 #PO_4^(-3)# = 1 x 2 = 2

Side on the right:

#Mg^(2+)# = 3 #NO_3^(-1)# = 1 x 6 = 6 #K^(1+)# = 1 x 6 = 6 #PO_4^(-3)# = 2

Last Response:

#color(green)3Mg(NO_3)_2# + #color(red)2K_3PO_4# #rarr# #Mg_3(PO_4)_2# + #color(blue)6KNO_3# (balanced)

Owen Ambrose · Answer

undefined The balanced equation is: 
3 Mg(NO3)2 + 2 K3PO4 → Mg3(PO4)2 + 6 KNO3

How do you balance #Mg(NO_3)_2 + K_3PO_4 -&gt; Mg_3(PO_4)_2 + KNO_3#?

How do you balance #Mg(NO_3)_2 + K_3PO_4 -> Mg_3(PO_4)_2 + KNO_3#?