1. Determine the correct formulas for all the reactants and products in the reaction.
2. Begin balancing with the most complicated-looking group. A polyatomic ion that appears unchanged on both sides of the equation can be counted as a single unit.
3. Save the elemental (single elements) reactant and products for last, especially if it is hydrogen or oxygen. Keep your eye out for diatomic molecules such as oxygen, hydrogen, and the halogens.
4. If you get stuck, double the most complicated-looking group and try again.
5. Finally, make sure that all coefficients are in the lowest-possible ratio.
6. Know when to quit! None of the reactions you will encounter will be that difficult. If the coefficients are getting wild, double-check what you’ve done since you may have a simple mistake.

    When balancing reactions, keep your hands off the subscripts! Use only coefficients to balance chemical equations. Now let’s try an example. When you solve it yourself, make sure to follow the steps!

 

 

    Write the balanced equation for the reaction between chlorine and sodium bromide, which produces bromine and sodium chloride.

 

 

    First write the chemical formulas—be on the lookout for the diatomic elements (such as Cl₂):

    Next, find the reagent with the scariest subscripts. In this case, start with Cl₂. You need a coefficient of 2 in front of NaCl, which then requires a coefficient of 2 in front of NaBr. The balanced equation becomes

 

 

    Finally, count up everything to make sure you balanced the equation correctly. You have two chlorine atoms, two sodium atoms, and two bromines on the reactants side and two bromines, two sodiums, and two chlorines on the products side. You’re done.

 

 

    Write the balanced equation for the reaction between aluminum sulfate and calcium chloride, which produces aluminum chloride and calcium sulfate.

 

 

    Write the chemical formulas on their correct sides:

    In this reaction, the aluminum sulfate looks the most complicated, so start there. Look at what happens with sulfate—since it remains sulfate on the right side of the reaction, treat it as a unit. You have three on the left side and only one on the right side, so place a coefficient of 3 in front of calcium sulfate. Now deal with the aluminum. You have three on the left and one on the right, so place a coefficient of 2 in front of aluminum chloride. Last, you must place a coefficient of 3 in front of calcium chloride.

 

 

    Count the atoms on both sides of the reaction and you’ll see that you’re done