How do you balance #Mg + H_3PO_4 -> Mg_3(PO_4)_2 + H_2#?

1 Answer
May 11, 2018

Answer:

#2H_3PO_4 -> Mg_3(PO_4)_2#

Explanation:

Balancing chemical equations always involves a bit of algebra, and maybe some trial-and-error. The objective is always to ensure that the numbers (moles) of each element are the same on each side of the equation.

In this case it is a redox reaction, so sometimes the use of the oxidation state changes can help. First, the phosphate ion is not changing, just switching partners. The magnesium is being oxidized (#Mg^0 -> Mg^(2+)#) and the hydrogen is being reduced (#H^1 -> H^0#). If we balance that charge transfer, we balance the element quantities involved.

#Mg + H_3PO_4 → Mg_3(PO_4)_2 + H_2#

The key changes are:
#Mg^0 → Mg^(+2) xx 3# and
#H^(1) xx 3→ H^(0) xx 2#

#Mg^0 → Mg^(+2)# requires giving up 2 electrons
#H^(1)→ H^(0)# requires adding 1 electron

So, there must be twice as many #H# atoms involved as #Mg# atoms. Balancing the #H# first requires increasing the left side by 2 and the right side by 3:
#2H^(1) xx 3→ 3H^(0) xx 2# 6 electrons transfered.

That means the #PO_4# ion is also changed on the left, and balanced on the right:
#2H_3PO_4 -> Mg_3(PO_4)_2#

Put them all together and check the #Mg# balance:
#3Mg + 2H_3PO_4 → Mg_3(PO_4)_2 + 3H_2#

Total Balance:
#" " Left" "Right#
#Mg" 3" " "3#
#H" " 6 " "6#
#PO_4" 2" " "2#