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

##### 1 Answer

#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" + "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

Next, we start of by balancing the elements that have the least occurrence.

There is 1

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

There are 3

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

There are 2

#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

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

Remember to put back the state symbols!