We will use example 12 from the solubility lab as an example. I hope by now that you have looked at all the links for “Help on writing formulas” and “Help on using the Solubility rules”. Lets take this in steps:

Place the yield sign to the right of the reactants

12. K₂CrO₄ + Fe(NO₃)₃ ®

Write out correct formulas of the reactants. You can find information on writing reactants from the link on “Help on writing formulas”

12. K₂CrO₄ + Fe(NO₃)₃ ® KNO₃ + FeCrO₄

Replace the cation and anions of the reactants without moving subscripts over, indicated in blue. (Do not remove subscripts of polyatomic ions)

12. K₂CrO₄ + Fe(NO₃)₃ ® KNO₃ + Fe₂(CrO₄)₃

Determine the charge of the cation and anions and make sure they equal zero. Notice that Fe is Fe⁺³ and CrO₄ is CrO₄^-2 therefore you need two irons for three chromates. The way I new the iron was Fe⁺³ was because in the reactant each NO3 is –1 which make the has to make the iron Fe⁺³. Remember iron is a type II ion.

Now that we have the correct formulas for the products the equation can be balanced. The equation can only be balanced when the all the formulas charges equal zero.

For information on balancing equation visit the link on “Help in Balancing equations”.

12. 3 K₂CrO₄ + 2 Fe(NO₃)₃ ® 6 KNO₃ + Fe₂(CrO₄)₃

The last step is to place the appropriate state of matter for each reactant and products. Since the reactant were both solution being mixed we can place an (aq) symbol to the right of each. The products on the other hand is a little tougher. Here we need the solubility rules. For more information on the solubility rules go the link on “help using the solubility rules”.

12. 3 K₂CrO₄ (aq) + 2 Fe(NO₃)₃ (aq) ® 6 KNO₃ (aq) + Fe₂(CrO₄)₃ (s)

From the rules we find that the only product that will precipitate out into a solid is the Fe₂(CrO₄)₃.