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(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
Right side:
#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
Right side:
#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
Right side:
#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
Right side:
#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
Right side:
#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
Right side:
#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
Right side:
#Mg^(2+)# = 3
#NO_3^(-1)# = 1 x 6 = 6
#K^(1+)# = 1 x 6 = 6
#PO_4^(-3)# = 2
Final Answer:
#color(green)3Mg(NO_3)_2# + #color(red)2K_3PO_4# #rarr# #Mg_3(PO_4)_2# + #color(blue)6KNO_3# (balanced)
