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

May 11, 2018

$2 {H}_{3} P {O}_{4} \to M {g}_{3} {\left(P {O}_{4}\right)}_{2}$

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 ($M {g}^{0} \to M {g}^{2 +}$) and the hydrogen is being reduced (${H}^{1} \to {H}^{0}$). 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 $M g$ 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 $P {O}_{4}$ ion is also changed on the left, and balanced on the right:
$2 {H}_{3} P {O}_{4} \to M {g}_{3} {\left(P {O}_{4}\right)}_{2}$

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

Total Balance:
$\text{ " Left" } R i g h t$
$M g \text{ 3" " } 3$
$H \text{ " 6 " } 6$
$P {O}_{4} \text{ 2" " } 2$