Question

Question #fee47

Answer 1

You balance an equation with parentheses the same way as with other equations.

You must remember to multiply the subscripts of atoms inside the parentheses by the subscripts outside the parentheses.

Explanation:

For example, balance the equation:

`"Fe(NO"_3)_2 + "Na"_3"PO"_4 → "Fe"_3("PO"_4)_2 + "NaNO"_3`

The standard method is to balance by inspection:

• First balance all atoms other than `"O"` and `"H"`.
• Then balance `"O"`.
• Then balance `"H"`.

Start with the most complicated formula. Here, it is `"Fe"_3("PO"_4)_2`:

We have `"three Fe"` atoms on the right, so we need to have `"three Fe"` atoms on the left hand side of the equation.

`color(red)(3)"Fe(NO"_3)_2 + "Na"_3"PO"_4 → color(red)(1)"Fe"_3("PO"_4)_2 + "NaNO"_3`

We have `"two P"` atoms on the right side (one `"P"` atom, which is inside the parentheses, is multiplied by the 2 outside). So we need `"two P"` atoms on the left hand side of the equation.

`color(red)(3)"Fe(NO"_3)_2 + color(teal)(2)"Na"_3"PO"_4 → color(red)(1)"Fe"_3("PO"_4)_2 + "NaNO"_3`

Now we have `"six N"` atoms on the left hand side (1 `"N"` atom inside the parentheses is multiplied by the 2 outside and again by the 3 in front of the whole formula). We need `"six N"` atoms on the right hand side of the equation.

`color(red)(3)"Fe(NO"_3)_2 + color(teal)(2)"Na"_3"PO"_4 → color(red)(1)"Fe"_3("PO"_4)_2 + color(blue)(6)"NaNO"_3`

We have `"6 Na"` atoms on the left and `"6 Na"` atoms on the right, so `"Na"` is balanced.

Now we want to balance `"O"` atoms. The total number of `"O"` atoms on the left hand side of the equation is 26. So we have 8 + 18 = 26 `"O"` atoms on the right. `"O"` atoms are balanced.

The balanced equation is:

`3"Fe(NO"_3)_2 + 2"Na"_3"PO"_4 → "Fe"_3("PO"_4)_2 + 6"NaNO"_3`

Answer 2

We balance by inspection, but we treat each group inside parentheses as if it were a single atom.

This is sometimes called the "EASY" way.

Explanation:

For example, balance the equation:

`"Fe(NO"_3)_2 + "Na"_3"PO"_4 → "Fe"_3("PO"_4)_2 + "NaNO"_3`

The groups inside parentheses behave as a unit, so we treat the `"NO"_3` as if it were an atom `"X"`, and `"PO"_4` at if it were an atom `"Y"`.

Then the equation becomes

`"FeX"_2 + "Na"_3"Y" → "Fe"_3"Y"_2 + "NaX"`

We begin with `"Fe"_3"Y"_2` and balance `"Fe"`.

`color(red)(3)"FeX"_2 + "Na"_3"Y" → color(red)(1)"Fe"_3"Y"_2 + "NaX"`

We have `"2 Y"` groups on the right hand side, so we need `"2 Y"` groups on the left hand side of the equation.

`color(red)(3)"FeX"_2 + color(teal)(2)"Na"_3"Y" → color(red)(1)"Fe"_3"Y"_2 + "NaX"`

We have `"6 X"` groups on the left hand side. We need `"6 X"` groups on the right hand side of the equation.

`color(red)(3)"FeX"_2 + color(teal)(2)"Na"_3"Y" → color(red)(1)"Fe"_3"Y"_2 + color(blue)(6)"NaX"`

All atoms and groups are now balanced.

The balanced equation is:

`color(red)(3"FeX"_2 + 2"Na"_3"Y" → "Fe"_3"Y"_2 + 6"NaX")`

Now all we have to do is re-insert the original groups represented by `"X"` and `"Y"`. We get

`"3Fe(NO"_3)_2 + "2Na"_3"PO"_4 → "Fe"_3("PO"_4)_2 + "6NaNO"_3`