Question

How do you balance `Na(s) + NaNO_3(aq) -> Na_2O(aq) + N_2(g)`?

Answer 1

`10"Na"(s) + 2"NaNO"_3(aq) -> 6"Na"_2"O"(aq) + "N"_2(g)`

Explanation:

After balancing the equation, you should see that the number of `"Na"`, `"O"` and `"N"` should be the same on both sides of the equation. This is the original equation.

`"Na" + "NaNO"_3 -> "Na"_2"O" + "N"_2`

When balancing the equations, I omit the state symbols. I will add them back in the final balanced equation.

The trick to balancing the equation is to fix the compound with the most elements present to a ratio of `1`. In this case, it is `"NaNO"_3`, as it contains all 3 elements.

Next, we start of by balancing the elements that have the least occurrence. `"Na"` appeared in 3 times, while `"N"` and `"O"` appeared only 2 times, so we start with `"N"` first.

`color(green)"Balancing N"`

There is 1 `"N"` on the left-hand side (LHS) and 2 `"N"` on the right-hand side (RHS). To balance `"N"`, we only need half the amount of `"N"_2` on the RHS.

`"Na" + "NaNO"_3 -> "Na"_2"O" + 1/2 "N"_2`

`color(green)"Balancing O"`

There are 3 `"O"` on the LHS and 1 `"O"` on the RHS. To balance `"O"`, we need 3 times the amount of `"Na"_2"O"` on the RHS.

`"Na" + "NaNO"_3 -> 3"Na"_2"O" + 1/2 "N"_2`

`color(green)"Balancing Na"`

There are 2 `"Na"` on the LHS and 6 `"Na"` on the RHS. Both the number of `"NaNO"_3` and `"Na"_2"O"` are fixed already, which means that we can only change the number of `"Na"`. To balance `"Na"`, we need 5 more `"Na"` on the LHS.

`5"Na" + "NaNO"_3 -> 3"Na"_2"O" + 1/2 "N"_2`

The balancing is now complete. However, some people do not like fractions as coefficients. So to get rid of the `1/2`, multiply everything by `2`. It becomes

`10"Na" + 2"NaNO"_3 -> 6"Na"_2"O" + "N"_2`

Remember to put back the state symbols!