# What is the balanced chemical equation of Li(s) + H3PO4(aq) ?

Oct 18, 2015

$6 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

#### Explanation:

Lithium metal will react with phosphoric acid to produce lithium phosphate, ${\text{Li"_3"PO}}_{4}$, and hydrogen gas, ${\text{H}}_{2}$.

${\text{Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

To balance this chemical equation, notice that the only atoms that are not balanced are lithium and hydrogen.

More specifically, you have one lithium atom on the reactants' side, and three on the products' side. This means that you need to multiply the former by $3$

$3 {\text{Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

The only atoms that remain unbalanced are the hydrogen atoms. This time you have three of these on the reactants's side, and only two on the products' side.

This implies that you're going to have to multiply the hydrogen atoms on the reactants' side by $2$ and those on the products' side by $3$ to get a total of six atoms on each side.

$3 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

Now you've unbalanced the number of phosphorus and oxygen atoms. To fix this, you can treat the phosphate anion, ${\text{PO}}_{4}^{3 -}$, as a unit.

This means that you will need to multiply the lithium phosphate by $2$ to get

$3 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

Finally, balance the lithium atoms again by multiplying the lithium metal by $6$ instead of by $3$.

$6 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

And now the chemical equation is balanced.