Question

Balance the equation using the

correct coefficient:
lithium + water produces lithium
hydroxide + hydrogen gas
[ ? ]Li + 2H2O → LiOH + H2
Enter

Answer 1

`2\; {\rm Li} + 2\; {\rm H_(2) O} \to 2\; {\rm LiOH} + 1\; {\rm H_(2)}`.

Step-by-step explanation:

Assume that the coefficient of
`{\rm LiOH}` is
`1`:

`?\; {\rm Li} + ?\; {\rm H_(2) O} \to 1\; {\rm LiOH} + ?\; {\rm H_(2)}`.

The product side of this equation would include:

• 
  `1`
  `{\rm Li` atom.
• 
  `1`
  `{\rm O}` atom.
• The number of
  `{\rm H}` atoms on the product side is not known since the coefficient of
  `{\rm H_(2)}` is not yet found.

By the conservation of atoms, the reactant side of this equation should also include:

• 
  `1`
  `{\rm Li}` atom, and
• 
  `1`
  `{\rm O}` atom.

The only reactant that includes
`{\rm Li}` atoms is
`{\rm Li}\!`. Thus, the coefficient of the reactant
`\!{\rm Li}` should be
`1` to ensure that the reactant side of this equation includes exactly
`1\!`
`{\rm Li}\!\!` atom.

Likewise, the only reactant that includes
`{\rm O}` atoms is
`{\rm H_(2)O}`, with one
`{\rm O}\!` atom in each formula unit of
`{\rm H_(2)O}\!`. There needs to be
`1\!`
`\!{\rm O}` atom on the reactant side of this equation. Therefore, the coefficient of the reactant
`{\rm H_(2)O}\!\!` should also be
`1`:

`1\; {\rm Li} + 1\; {\rm H_(2) O} \to 1\; {\rm LiOH} + ?\; {\rm H_(2)}`.

The reactant side of this equation now include:

• 
  `1`
  `{\rm Li}` atom.
• 
  `1`
  `{\rm O}` atom.
• 
  `2`
  `{\rm H}` atoms.

By the conservation of atoms, the product side of this equation should also include:

• 
  `1`
  `{\rm Li}` atom.
• 
  `1`
  `{\rm O}` atom.
• 
  `2`
  `{\rm H}` atoms.

The product
`1\; {\rm LiOH}` alone would account for
`1\; {\rm Li}` atom,
`1\; {\rm O}` atom, and
`1\; {\rm H}` atom. Thus, there would be only
`1` more
`{\rm H}` atom for the other product,
`{\rm H_(2)}`. However, since there are
`2`
`{\rm H}\!` atoms in every formula unit of
`{\rm H_(2)}\!`, the coefficient of
`\!{\rm H_(2)}` would need to be
`(1/2)`:

`1\; {\rm Li} + 1\; {\rm H_(2) O} \to 1\; {\rm LiOH} + (1/2)\; {\rm H_(2)}`.

Atoms of each element are indeed conserved in this equation.

Eliminate the fractions by multiplying all coefficients by
`2` (least common denominator of all coefficients in this equation):

`2\; {\rm Li} + 2\; {\rm H_(2) O} \to 2\; {\rm LiOH} + 1\; {\rm H_(2)}`.