Balance the following chemical equation:

[2, 1, 2]
`\qquad` SOLUTION[0]`\text{H}_2 +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{H}_2\text{O}`

There are `2 \text{ O}` on the left and only `1` on the right, so multiply `\text{H}_2\text{O}` by `\blue{2}`.

`\qquad \text{H}_2 + \text{O}_2 \rightarrow \blue{2}\text{H}_2\text{O}`

That gives us `4 \text{ H}` on the right and only `2` on the left, so multiply `\text{H}_2` by `\red{2}`.

`\qquad \red{2}\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}`

The balanced equation is:

`\qquad 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}`

[1, 4, 1, 4]
`\qquad` SOLUTION[0]`\text{CH}_4 +` SOLUTION[1]`\text{Cl}_2 \rightarrow` SOLUTION[2]`\text{CCl}_4 +` SOLUTION[3]`\text{HCl}`

`\text{C}` is already balanced.

There are `4 \text{ H}` on the left and only `1` on the right, so multiply `\text{HCl}` by `\blue{4}`.

`\qquad \text{CH}_4 + \text{Cl}_2 \rightarrow \text{CCl}_4 + \blue{4}\text{HCl}`

That gives us `8 \text{ Cl}` on the right and only `2` on the left, so multiply `\text{Cl}_2` by `\red{4}`.

`\qquad \text{CH}_4 + \red{4}\text{Cl}_2 \rightarrow \text{CCl}_4 + 4\text{HCl}`

The balanced equation is:

`\qquad \text{CH}_4 + 4\text{Cl}_2 \rightarrow \text{CCl}_4 + 4\text{HCl}`

[4, 3, 2]
`\qquad` SOLUTION[0]`\text{Al} +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{Al}_2\text{O}_3`

There are `2 \text{ O}` on the left and `3` on the right. The lowest common denominator is `6`, so multiply `\text{O}_2` by `\blue{3}` and `\text{Al}_2\text{O}_3` by `\red{2}`.

`\qquad \text{Al} + \blue{3}\text{O}_2 \rightarrow \red{2}\text{Al}_2\text{O}_3`

That gives us `4 \text{ Al}` on the right and only `1` on the left, so multiply `\text{Al}` by `\pink{4}`.

`\qquad \pink{4}\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3`

The balanced equation is:

`\qquad 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3`

[1, 2, 1, 2]
`\qquad` SOLUTION[0]`\text{CH}_4 +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{CO}_2 +` SOLUTION[3]`\text{H}_2\text{O}`

`\text{C}` is already balanced.

There are `4 \text{ H}` on the left and `2` on the right, so multiply `\text{H}_2\text{O}` by `\blue{2}`.

`\qquad \text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \blue{2}\text{H}_2\text{O}`

That gives us `4 \text{ O}` on the right and only `2` on the left, so multiply `\text{O}_2` by `\red{2}`. (Since oxygen is by itself on the left, it should be done at the end because you can give it a coefficient without affecting another element.)

`\qquad \text{CH}_4 + \red{2}\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}`

The balanced equation is:

`\qquad \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}`

[2, 1, 2, 1]
`\qquad` SOLUTION[0]`\text{NaBr} +` SOLUTION[1]`\text{Cl}_2 \rightarrow` SOLUTION[2]`\text{NaCl} +` SOLUTION[3]`\text{Br}_2`

There is `1 \text{ Br}` on the left and `2` on the right, so multiply `\text{NaBr}` by `\blue{2}`.

`\qquad \blue{2}\text{NaBr} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{Br}_2`

There are `2 \text{ Cl}` on the left and `1` on the right, so multiply `\text{NaCl}` by `\red{2}`.

`\qquad 2\text{NaBr} + \text{Cl}_2 \rightarrow \red{2}\text{NaCl} + \text{Br}_2`

Now `\text{Na}` is balanced again.

The balanced equation is:

`\qquad 2\text{NaBr} + \text{Cl}_2 \rightarrow 2\text{NaCl} + \text{Br}_2`

[1, 2, 1, 1]
`\qquad` SOLUTION[0]`\text{Mg} +` SOLUTION[1]`\text{HCl} \rightarrow` SOLUTION[2]`\text{MgCl}_2 +` SOLUTION[3]`\text{H}_2`

There are `2 \text{ Cl}` on the right and only `1` on the left, so multiply `\text{HCl}` by `\blue{2}`.

`\qquad \text{Mg} + \blue{2}\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2`

Now all atoms are balanced; the balanced equation is:

`\qquad \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2`

[2, 2, 1, 4]
`\qquad` SOLUTION[0]`\text{NH}_4\text{NO}_3 \rightarrow` SOLUTION[1]`\text{N}_2 +` SOLUTION[2]`\text{O}_2 +` SOLUTION[3]`\text{H}_2\text{O}`

Start with the compound that has the most elements `(\text{NH}_4\text{NO}_3)`.

There are `2 \text{N}` on the left and `2 \text{N}` on the right, so `\text{N}` is already balanced.

There are `4 \text{ H}` on the left and `2` on the right, so multiply `\text{H}_2\text{O}` by `\blue{2}`.

`\qquad \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2 + \text{O}_2 + \blue{2}\text{H}_2\text{O}`

That gives us `3 \text{ O}` on the left and `4` on the right. If we try giving `\text{O}_2` a coefficient of `\red{\frac{1}{2}}`, it gives us `3 \text{ O}` on both sides.

`\qquad \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2 + \red{\frac{1}{2}}\text{O}_2 + 2\text{H}_2\text{O}`

Since fractions are not usually used as coefficients, multiply everything by `2` to get rid of the fraction.

`\qquad 2\text{NH}_4\text{NO}_3 \rightarrow 2\text{N}_2 + 1\text{O}_2 + 4\text{H}_2\text{O}`

The balanced equation is:

`\qquad 2\text{NH}_4\text{NO}_3 \rightarrow 2\text{N}_2 + \text{O}_2 + 4\text{H}_2\text{O}`

[1, 3, 2, 3]
`\qquad` SOLUTION[0]`\text{C}_2\text{H}_6\text{O} +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{CO}_2 +` SOLUTION[3]`\text{H}_2\text{O}`

For a combustion reaction, it is usually easiest to start with `\text{C}`.

There are `2 \text{ C}` on the left and `1` on the right, so multiply `\text{CO}_2` by `\blue{2}`.

`\qquad \text{C}_2\text{H}_6\text{O} + \text{O}_2 \rightarrow \blue{2}\text{CO}_2 + \text{H}_2\text{O}`

There are `6 \text{ H}` on the left and `2` on the right, so multiply `\text{H}_2\text{O}` by `\red{3}`.

`\qquad \text{C}_2\text{H}_6\text{O} + \text{O}_2 \rightarrow 2\text{CO}_2 + \red{3}\text{H}_2\text{O}`

That gives us `7 \text{ O}` on the right and `3` on the left, so multiply `\text{O}_2` by `\pink{3}`.

`\qquad \text{C}_2\text{H}_6\text{O} + \pink{3}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}`

The balanced equation is:

`\qquad \text{C}_2\text{H}_6\text{O} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}`

[2, 1, 2]
`\qquad` SOLUTION[0]`\text{Mg} +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{MgO}`

There are `2 \text{ O}` on the left and only `1` on the right, so multiply `\text{MgO}` by `\blue{2}`.

`\qquad \text{Mg} + \text{O}_2 \rightarrow \blue{2}\text{MgO}`

That gives us `2 \text{ Mg}` on the right and only `1` on the left, so multiply `\text{Mg}` by `\red{2}`.

`\qquad \red{2}\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}`

The balanced equation is:

`\qquad 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}`

[2, 6, 2, 3]
`\qquad` SOLUTION[0]`\text{Al} +` SOLUTION[1]`\text{HCl} \rightarrow` SOLUTION[2]`\text{AlCl}_3 +` SOLUTION[3]`\text{H}_2`

There is `1 \text{ H}` and `1 \text{ Cl}` on the left and `3 \text{ Cl}` and `2 \text{ H}` on the right. The lowest common denominator for the right is `6`, so multiply `\text{AlCl}_3` by `\blue{2}` and `\text{H}_2` by `\red{3}`.

`\qquad \text{Al} + \text{HCl} \rightarrow \blue{2}\text{AlCl}_3 + \red{3}\text{H}_2`

That gives us `6 \text{ H}` and `6 \text{ Cl}` on the right, so multiply `\text{HCl}` by `\pink{6}`.

`\qquad \text{Al} + \pink{6}\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2`

That gives us `2 \text{ Al}` on the right and only `1` on the left, so multiply `\text{Al}` by `\green{2}`.

`\qquad \green{2}\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2`

The balanced equation is:

`\qquad 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2`

[3, 2, 1, 6]
`\qquad` SOLUTION[0]`\text{CaCl}_2 +` SOLUTION[1]`\text{Na}_3\text{PO}_4 \rightarrow` SOLUTION[2]`\text{Ca}_3\text{(PO}_4\text{)}_2 +` SOLUTION[3]`\text{NaCl}`

We can treat the phosphate polyatomic ion `\text{(PO}_4\text{)}` as an atom, symbolized by `\green{X}`:

`\qquad \text{CaCl}_2 + \text{Na}_3\green{\text{PO}_4} \rightarrow \text{Ca}_3(\green{\text{PO}_4\text{}})_2 + \text{NaCl}`

`\qquad \text{CaCl}_2 + \text{Na}_3\green{X} \rightarrow \text{Ca}_3\green{X}_2 + \text{NaCl}`

There is `1 \space X` on the left and `2 \space X` on the right, so multiply `\text{Na}_3X` by `\blue{2}`.

`\qquad \text{CaCl}_2 + \blue{2}\text{Na}_3X \rightarrow \text{Ca}_3X_2 + \text{NaCl}`

There are `6 \text{ Na}` on the left and only `1` on the right, so multiply `\text{NaCl}` by `\red{6}`.

`\qquad \text{CaCl}_2 + 2\text{Na}_3X \rightarrow \text{Ca}_3X_2 + \red{6}\text{NaCl}`

That gives us `6 \text{ Cl}` on the right and only `2` on the left, so multiply `\text{CaCl}_2` by `\pink{3}`.

`\qquad \pink{3}\text{CaCl}_2 + 2\text{Na}_3X \rightarrow \text{Ca}_3X_2 + 6\text{NaCl}`

Now `\text{Ca}` is balanced too.

Replacing `\text{PO}_4` for `X`, the balanced equation is:

`\qquad 3\text{CaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ca}_3\text{(PO}_4\text{)}_2 + 6\text{NaCl}`

[1, 3, 2, 2]
`\qquad` SOLUTION[0]`\text{N}_2\text{H}_4 +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{NO}_2 +` SOLUTION[3]`\text{H}_2\text{O}`

There are `2 \text{ N}` on the left and only `1` on the right, so multiply `\text{NO}_2` by `\blue{2}`.

`\qquad \text{N}_2\text{H}_4 + \text{O}_2 \rightarrow \blue{2}\text{NO}_2 + \text{H}_2\text{O}`

There are `4 \text{ H}` on the left and only `2` on the right, so multiply `\text{H}_2\text{O}` by `\red{2}`.

`\qquad \text{N}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{NO}_2 + \red{2}\text{H}_2\text{O}`

That gives us `6 \text{ O}` on the right and only `2` on the left, so multiply `\text{O}_2` by `\pink{3}`. (Since oxygen is by itself on the left, it should be done at the end because you can give it a coefficient without affecting another element.)

`\qquad \text{N}_2\text{H}_4 + \pink{3}\text{O}_2 \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O}`

The balanced equation is:

`\qquad \text{N}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O}`

[3, 4, 1, 4]
`\qquad` SOLUTION[0]`\text{Fe} +` SOLUTION[1]`\text{H}_2\text{O} \rightarrow` SOLUTION[2]`\text{Fe}_3\text{O}_4 +` SOLUTION[3]`\text{H}_2`

There are `3 \text{ Fe}` on the right and only `1` on the left, so multiply `\text{Fe}` by `\blue{3}`.

`\qquad \blue{3}\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2`

There are `4 \text{ O}` on the right and only `1` on the left, so multiply `\text{H}_2\text{O}` by `\red{4}`.

`\qquad 3\text{Fe} + \red{4}\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2`

That gives us `8 \text{ H}` on the left and only `2` on the right, so multiply `\text{H}_2` by `\pink{4}`. (Since hydrogen is by itself on the right, it should be done at the end because you can give it a coefficient without affecting another element.)

`\qquad 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \pink{4}\text{H}_2`

The balanced equation is:

`\qquad 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2`

[1, 1, 2]
`\qquad` SOLUTION[0]`\text{NH}_4\text{NO}_3 \rightarrow` SOLUTION[1]`\text{N}_2\text{O} +` SOLUTION[2]`\text{H}_2\text{O}`

There are `2 \text{ N}` on the right and `2` on the left, so `\text{N}` is already balanced.

There are `4 \text{ H}` on the left and only `2` on the right, so multiply `\text{H}_2\text{O}` by `\blue{2}`.

`\qquad \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + \blue{2}\text{H}_2\text{O}`

There are `3 \text{ O}` on the right and `3` on the left, so `\text{O}` is already balanced.

The balanced equation is:

`\qquad \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O}`

[2, 2, 1]
`\qquad` SOLUTION[0]`\text{HgO} \rightarrow` SOLUTION[1]`\text{Hg} +` SOLUTION[2]`\text{O}_2`

There are `2 \text{ O}` on the right and only `1` on the left, so multiply `\text{HgO}` by `\blue{2}`.

`\qquad \blue{2}\text{HgO} \rightarrow \text{Hg} + \text{O}_2`

Now there are `2 \text{ Hg}` on the left and only `1` on the right, so multiply `\text{Hg}` by `\red{2}`.

`\qquad 2\text{HgO} \rightarrow \red{2}\text{Hg} + \text{O}_2`

The balanced equation is:

`\qquad 2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2`

[1, 4, 1, 2]
`\qquad` SOLUTION[0]`\text{SiO}_2 +` SOLUTION[1]`\text{HF} \rightarrow` SOLUTION[2]`\text{SiF}_4 +` SOLUTION[3]`\text{H}_2\text{O}`

There is `1 \text{ Si}` on the right and `1` on the left, so `\text{Si}` is already balanced.

There are `2 \text{ O}` on the left and only `1` on the right, so multiply `\text{H}_2\text{O}` by `\blue{2}`.

`\qquad \text{SiO}_2 + \text{HF} \rightarrow \text{SiF}_4 + \blue{2}\text{H}_2\text{O}`

Now there are `4 \text{ H}` on the right and only `1` on the left, so multiply `\text{HF}` by `\red{4}`.

`\qquad \text{SiO}_2 + \red{4}\text{HF} \rightarrow \text{SiF}_4 + 2\text{H}_2\text{O}`

Now `\text{F}` is balanced too.

The balanced equation is:

`\qquad \text{SiO}_2 + 4\text{HF} \rightarrow \text{SiF}_4 + 2\text{H}_2\text{O}`

[1, 2, 1, 2]
`\qquad` SOLUTION[0]`\text{Mg(OH)}_2 +` SOLUTION[1]`\text{HCl} \rightarrow` SOLUTION[2]`\text{MgCl}_2 +` SOLUTION[3]`\text{H}_2\text{O}`

There is `1 \text{ Mg}` on the right and `1` on the left, so `\text{Mg}` is already balanced.

There are `2 \text{ Cl}` on the right and only `1` on the left, so multiply `\text{HCl}` by `\blue{2}`.

`\qquad \text{Mg(OH)}_2 + \blue{2}\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O}`

Now there are `4 \text{ H}` on the left and only `2` on the right, so multiply `\text{H}_2\text{O}` by `\red{2}`.

`\qquad \text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \red{2}\text{H}_2\text{O}`

Now `\text{O}` is balanced too.

The balanced equation is:

`\qquad \text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}`

[2, 1, 1, 4]
`\qquad` SOLUTION[0]`\text{H}_2\text{SO}_4 +` SOLUTION[1]`\text{Pb(OH)}_4 \rightarrow` SOLUTION[2]`\text{Pb(SO}_4\text{)}_2 +` SOLUTION[3]`\text{H}_2\text{O}`

We can treat the sulfate polyatomic ion `\text{(SO}_4\text{)}` as an atom, symbolized by `\green{X}`:

`\qquad \text{H}_2\green{\text{SO}_4} + \text{Pb(OH)}_4 \rightarrow \text{Pb(}\green{\text{SO}_4}\text{)}_2 + \text{H}_2\text{O}`

`\qquad \text{H}_2\green{X} + \text{Pb(OH)}_4 \rightarrow \text{Pb}\green{X}_2 + \text{H}_2\text{O}`

There is `1 \space X` on the left and `2 \space X` on the right, so multiply `\text{H}_2X` by `\blue{2}`.

`\qquad \blue{2}\text{H}_2X + \text{Pb(OH)}_4 \rightarrow \text{Pb}X_2 + \text{H}_2\text{O}`

That gives us `8 \text{ H}` on the left and only `2` on the right, so multiply `\text{H}_2\text{O}` by `\red{4}`.

`\qquad 2\text{H}_2X + \text{Pb(OH)}_4 \rightarrow \text{Pb}X_2 + \red{4}\text{H}_2\text{O}`

Everything is now balanced. Replacing `\text{SO}_4` for `X`, the balanced equation is:

`\qquad 2\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(}\text{SO}_4\text{)}_2 + 4\text{H}_2\text{O}`

[1, 9, 1, 6]
`\qquad` SOLUTION[0]`\text{As}_4\text{S}_6 +` SOLUTION[1]`\text{O}_2 \rightarrow` SOLUTION[2]`\text{As}_4\text{O}_6 +` SOLUTION[3]`\text{SO}_2`

`\text{As}` is already balanced.

There are `6 \text{ S}` on the left and only `1` on the right, so multiply `\text{SO}_2` by `\blue{6}`.

`\qquad \text{As}_4\text{S}_6 + \text{O}_2 \rightarrow \text{As}_4\text{O}_6 + \blue{6}\text{SO}_2`

That gives us `18 \text{ O}` on the right and only `2` on the left, so multiply `\text{O}_2` by `\red{9}`. (Since oxygen is by itself on the left, it should be done at the end because you can give it a coefficient without affecting another element.)

`\qquad \text{As}_4\text{S}_6 + \red{9}\text{O}_2 \rightarrow \text{As}_4\text{O}_6 + 6\text{SO}_2`

The balanced equation is:

`\qquad \text{As}_4\text{S}_6 + 9\text{O}_2 \rightarrow \text{As}_4\text{O}_6 + 6\text{SO}_2`

[1, 3, 2, 3]
`\qquad` SOLUTION[0]`\text{Cr}_2\text{O}_3 +` SOLUTION[1]`\text{Mg} \rightarrow` SOLUTION[2]`\text{Cr} +` SOLUTION[3]`\text{MgO}`

There are `2 \text{ Cr}` on the left and only `1` on the right, so multiply `\text{Cr}` by `\blue{2}`.

`\qquad \text{Cr}_2\text{O}_3 + \text{Mg} \rightarrow \blue{2}\text{Cr} + \text{MgO}`

There are `3 \text{ O}` on the left and only `1` on the right, so multiply `\text{MgO}` by `\red{3}`.

`\qquad \text{Cr}_2\text{O}_3 + \text{Mg} \rightarrow 2\text{Cr} + \red{3}\text{MgO}`

That gives us `3 \text{ Mg}` on the right and only `1` on the left, so multiply `\text{Mg}` by `\pink{3}`.

`\qquad \text{Cr}_2\text{O}_3 + \pink{3}\text{Mg} \rightarrow 2\text{Cr} + 3\text{MgO}`

The balanced equation is:

`\qquad \text{Cr}_2\text{O}_3 + 3\text{Mg} \rightarrow 2\text{Cr} + 3\text{MgO}`