How would you balance H3PO4 +Mg(OH)2-->Mg3(PO4)2 +H20?

1 Answer
Nov 1, 2015

Answer:

#2H_3PO_4# + #3Mg(OH)_2# = #Mg_3(PO_4)_2# + #6H_2O#

Explanation:

I can either tally all the atoms one by one or use my knowledge of ionic bonds to make the tally system much simpler.

I'll do the simpler version.

First re-write the chemical equation to avoid confusing myself:

#H_3PO_4# + #Mg(OH)_2# = #Mg_3(PO_4)_2# + #H#-#OH#

Notice that I had shown the line structure between the #H^+# and #OH^-# ions. Now, let's tally based on the subscripts.

left side:

#H^+# = 3

#PO_4^"3-"# = 1

#Mg^"2+"# = 1

#OH^-# = 2

right side:

#H^+# = 1

#PO_4^"3-"# = 2

#Mg^"2+"# = 3

#OH^-# = 1

Let's start balancing the largest ion, #PO_4^"3-"#.

#color (red) 2 H_3PO_4# + #Mg(OH)_2# = #Mg_3(PO_4)_2# + #H#-#OH#

left side:

#H^+# = 3 x #color (red) 2# = 6

#PO_4^"3-"# = 1 x #color (red) 2# = 2

#Mg^"2+"# = 1

#OH^-# = 2

right side:

#H^+# = 1

#PO_4^"3-"# = 2

#Mg^"2+"# = 3

#OH^-# = 1

Do not forget that since #H_3PO_4# is a substance, you need to also multiply the coefficient with the number of #H# atoms.

Next,

#2 H_3PO_4# + #color (green) 3Mg(OH)_2# = #Mg_3(PO_4)_2# + #H#-#OH#

left side:

#H^+# = 3 x 2 = 6

#PO_4^"3-"# = 1 x 2 = 2

#Mg^"2+"# = 1 x #color (green) 3# = 3

#OH^-# = 2 x #color (green) 3# = 6

right side:

#H^+# = 1

#PO_4^"3-"# = 2

#Mg^"2+"# = 3

#OH^-# = 1

Again, since #Mg(OH)_2# is a substance, whatever coefficient you apply to #Mg^"2+"# ion will also be applied to #OH^-# ion.

#2 H_3PO_4# + #3Mg(OH)_2# = #Mg_3(PO_4)_2# + #color (blue) 6##H#-#OH#

left side:

#H^+# = 3 x 2 = 6

#PO_4^"3-"# = 1 x 2 = 2

#Mg^"2+"# = 1 x 3 = 3

#OH^-# = 2 x 3 = 6

right side:

#H^+# = 1 x #color (blue) 6# = 6

#PO_4^"3-"# = 2

#Mg^"2+"# = 3

#OH^-# = 1 x #color (blue) 6# = 6

Reverting back to the original equation,

#2 H_3PO_4# + #3Mg(OH)_2# = #Mg_3(PO_4)_2# + #6H_2O#

The equation is now balanced.

Please take note this method is only applicable for IONIC bonding of polyatomic ions.