Question

What is the balanced chemical equation of $L{i_{(s)}} + {H_3}P{O_4}_{(aq)}$?

Answer 1

Hint :According to the Law of Conservation of Matter, matter cannot be produced or destroyed. Substances can change shape through a physical change, but their overall mass remains constant.

Complete Step By Step Answer:
According to the question, lithium reacts with phosphoric acid. As lithium metal reacts with phosphoric acid, lithium phosphate, $L{i_3}P{O_4}$, as well as hydrogen gas ${H_2}$ is formed.
$L{i_{(s)}} + {H_3}P{O_4}_{(aq)} \to L{i_3}P{O_4}_{(aq)} + {H_2}_{(g)} \uparrow $
Remember that the two atoms that are not balanced in this chemical equation are lithium and hydrogen.
There is one lithium atom on the reactants' side and three on the products' side. This means you must multiply the former by three.
$3L{i_{(s)}} + {H_3}P{O_4}_{(aq)} \to L{i_3}P{O_4}_{(aq)} + {H_2}_{(g)} \uparrow $
The only unbalanced atoms are the hydrogen atoms. Three of these are on the reactants' side and just two are on the products' side.
This means multiplying the hydrogen atoms on the reactants' side by two and the hydrogen atoms on the products' side by three to get a total of six atoms on each side.
$3L{i_{(s)}} + 2{H_3}P{O_4}_{(aq)} \to L{i_3}P{O_4}_{(aq)} + 3{H_2}_{(g)} \uparrow $
Now, the phosphorus and oxygen atoms have been shifted out of equilibrium. To correct this, consider the phosphate anion, $PO_4^{3 - }$, as a unit.
This suggests that you must multiply the lithium phosphate by two to obtain it.
$3L{i_{(s)}} + 2{H_3}P{O_4}_{(aq)} \to 2L{i_3}P{O_4}_{(aq)} + 3{H_2}_{(g)} \uparrow $
Now, multiply the lithium metal by $6$ instead of $3$ to rebalance the lithium atoms.
$6L{i_{(s)}} + 2{H_3}P{O_4}_{(aq)} \to 2L{i_3}P{O_4}_{(aq)} + 3{H_2}_{(g)} \uparrow $
Thus, the balanced equation of $L{i_{(s)}} + {H_3}P{O_4}_{(aq)}$is $6L{i_{(s)}} + 2{H_3}P{O_4}_{(aq)} \to 2L{i_3}P{O_4}_{(aq)} + 3{H_2}_{(g)} \uparrow $.

Note :
A balanced chemical equation has the same number of atoms on the reactant and component sides for each element involved in the reaction. This is a condition that the equation must meet in order to be consistent with the law of conservation of matter.