Stoichiometry

Step-by-Step

1. Write a balanced  equation for the reaction.

        Example

                        Potassium chlorate, when heated, decomposes to potassium chloride and oxygen.  How much potassium chloride and oxygen are produced using 50.0 g of potassium chlorate?

                               KClO₃    -------------->  KCl   +   O₂

                                                                        Δ _{delta means heated}

                              2KClO₃   ---------------> 2KCl  +  3O_2(g)

                                                      Δ

2. Write the equation again listing the known quantities from the problem. Use the variable x and y for the unknown quantities.

                                   50.0 g                            x g        y g

                               2KClO₃  --------------->  2KCl  + 3O_2(g)

3. Write the equation again with the formula weights beneath each known and unknown formula.

                                    50.0 g                            x g          y g

                               2KClO₃   ---------------> 2KCl  +  3O_{2 (g)}

                                              122.5 g/mol               74.5 g/mol    32 g/mol

Note: K = 19, Cl = 35.45, O =(16 x 3)                          O = (16 x 2)

4.  Examine the molar ratio.

                                                2:2:3 (the coefficients)

5.  Calculate the number of mol of the known formula.

                                50.0 g KClO₃  ÷  122.5 g/mol  =  0.408 mol KClO₃

6. Now, rewrite the equation with calculated number of mol.

                                50.0 g = 0.408 mol   

                                2KClO₃ ---------> 2KCl  +  3O₂

7. Examine the ratio of mol across the reaction.

                                 0.408 mol            0.408 mol   3/2 of 0.408 mol = 0.612 mol

                                2KClO₃ ---------> 2KCl  +  3O₂

8. Rewrite the equation showing molar quantities and formula weights.

                                  0.408 mol          0.408 mol  0.612 mol

                                2KClO₃  ---------> 2KCl  +  3O₂

                                              _{122.5 g/mol               74.5 g/mol          32 g/mol}

9.  Multiply the number of mol by each of their respective formula weights.

                        KCl = 0.408 mol x 74.5 g/mol  =  30.4 g

                        O₂ = 0.732 mol x 32 g/mol = 23.4 g

10. Notice that we only used the coefficients to obtain molar ratios! Never did we multiply the coefficient by the formula weight!

            DID WE SOLVE WHAT WAS ASKED FOR IN THE PROBLEM?

©William C. Willis 2011