Answer:
[tex]2\; {\rm Li} + 2\; {\rm H_{2} O} \to 2\; {\rm LiOH} + 1\; {\rm H_{2}}[/tex].
Explanation:
Assume that the coefficient of [tex]{\rm LiOH}[/tex] is [tex]1[/tex]:
[tex]?\; {\rm Li} + ?\; {\rm H_{2} O} \to 1\; {\rm LiOH} + ?\; {\rm H_{2}}[/tex].
The product side of this equation would include:
By the conservation of atoms, the reactant side of this equation should also include:
The only reactant that includes [tex]{\rm Li}[/tex] atoms is [tex]{\rm Li}\![/tex]. Thus, the coefficient of the reactant [tex]\!{\rm Li}[/tex] should be [tex]1[/tex] to ensure that the reactant side of this equation includes exactly [tex]1\![/tex] [tex]{\rm Li}\!\![/tex] atom.
Likewise, the only reactant that includes [tex]{\rm O}[/tex] atoms is [tex]{\rm H_{2}O}[/tex], with one [tex]{\rm O}\![/tex] atom in each formula unit of [tex]{\rm H_{2}O}\![/tex]. There needs to be [tex]1\![/tex] [tex]\!{\rm O}[/tex] atom on the reactant side of this equation. Therefore, the coefficient of the reactant [tex]{\rm H_{2}O}\!\![/tex] should also be [tex]1[/tex]:
[tex]1\; {\rm Li} + 1\; {\rm H_{2} O} \to 1\; {\rm LiOH} + ?\; {\rm H_{2}}[/tex].
The reactant side of this equation now include:
By the conservation of atoms, the product side of this equation should also include:
The product [tex]1\; {\rm LiOH}[/tex] alone would account for [tex]1\; {\rm Li}[/tex] atom, [tex]1\; {\rm O}[/tex] atom, and [tex]1\; {\rm H}[/tex] atom. Thus, there would be only [tex]1[/tex] more [tex]{\rm H}[/tex] atom for the other product, [tex]{\rm H_{2}}[/tex]. However, since there are [tex]2[/tex] [tex]{\rm H}\![/tex] atoms in every formula unit of [tex]{\rm H_{2}}\![/tex], the coefficient of [tex]\!{\rm H_{2}}[/tex] would need to be [tex](1/2)[/tex]:
[tex]1\; {\rm Li} + 1\; {\rm H_{2} O} \to 1\; {\rm LiOH} + (1/2)\; {\rm H_{2}}[/tex].
Atoms of each element are indeed conserved in this equation.
Eliminate the fractions by multiplying all coefficients by [tex]2[/tex] (least common denominator of all coefficients in this equation):
[tex]2\; {\rm Li} + 2\; {\rm H_{2} O} \to 2\; {\rm LiOH} + 1\; {\rm H_{2}}[/tex].
