Question

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

Answer 1

`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`