Example of chemical calculations:

How many grams of ferric oxide will be created if 27.9 grams of iron reacts along with water according to the following equation?

2Fe + 3H₂O → Fe₂O₃ + 3H₂↑

Solutions:

The equation weight of iron equals the gram atomic weight of iron times the number of atoms displays reacting in the equation, that is two.  Using Table 2:

Equation Weight Fe = 2× 55.8 grams

                                       = 111.6 grams

Since 27.9 g of iron react then the fraction of the equation weight which reacts is:

27.9 grams/ 111.6 grams = ¼

Therefore, 1/4 of the equation weight of ferric oxide will be created.

The equation weight of ferric oxide equals the gram molecular weight of ferric oxide times a number of molecules display formed in the equation that is one. Using Table 2:

Equation Weight Fe2O3 = 2(55.8 g) + 3(16.0 g)

                                            = 111.6 g + 48.0 g

                                            = 159.6 g

Therefore, the amount of ferric oxide formed is:

¼ (159.6 g) = 39.9 g