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

2 Answers
Mar 7, 2017

Answer:

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

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.

Mar 10, 2017

Answer:

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

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(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)