Example 4.1: Balancing Chemical Equations
Write a balanced equation for the reaction of molecular nitrogen
\(\ce{(N2)}\) and oxygen
\(\ce{(O2)}\) to form dinitrogen pentoxide.
Solution
First, write the unbalanced equation.
\(\ce{N2}\)\(\ce{ + }\)\(\ce{O2}\)\(\ce{->}\)\(\ce{N2O5}\)\(\ce{ }\) unbalanced
Next, count the number of each type of atom present in the unbalanced equation.
\(\mathrm{atoms}_{\mathrm{N,left}} = 2\)
\(\mathrm{atoms}_{\mathrm{N,right}} = 2\)
\(\mathrm{atoms}_{\mathrm{O,left}} = 2\)
\(\mathrm{atoms}_{\mathrm{O,right}} = 5\)
Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the
\(\ce{O2}\) and
\(\ce{N2O5}\) to integers that will yield
10 O atoms (the least common multiple for the O atom subscripts in these two formulas).
\(\ce{N2}\)\(\ce{ + }\)\(\ce{5O2}\)\(\ce{->}\)\(\ce{2N2O5}\)\(\ce{ }\)
\(\mathrm{atoms}_{\mathrm{N,left}} = 2\)
\(\mathrm{atoms}_{\mathrm{N,right}} = 2 \cdot 2\)
\(\mathrm{atoms}_{\mathrm{O,left}} = 5 \cdot 2\)
\(\mathrm{atoms}_{\mathrm{O,right}} = 2 \cdot 5\)
The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant
\(\ce{N2}\) to
2.
\(\ce{2N2}\)\(\ce{ + }\)\(\ce{5O2}\)\(\ce{->}\)\(\ce{2N2O5}\)\(\ce{ }\) balanced
\(\mathrm{atomsN}_{\mathrm{left}} = 2 \cdot 2\)
\(\mathrm{atomsN}_{\mathrm{right}} = 2 \cdot 2\)
\(\mathrm{atomsO}_{\mathrm{left}} = 5 \cdot 2\)
\(\mathrm{atomsO}_{\mathrm{right}} = 2 \cdot 5\)
The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.