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

1 Answer
Mar 5, 2016

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"#, #"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!