An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of iron(III) oxide will be formed upon the complete reaction of 32.2 grams of iron with excess oxygen gas ? iron(s) + oxygen(g) iron(III) oxide(s) moles iron(III) oxide

0.453 moles

Explanation:

The balanced equation for the reaction is:

2Fe(s) + 3O2(g)  ==>  2Fe2O3

From the equation,  mass of O2 involved = 16 x 2 x 3 = 96g

                                 mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

                                                                            = 100g

                Therefore 96g of O2 produced 100g of Fe2O3

                                  32.2g of O2 Will produce   100x32.2/96

                                                   = 33.54g of Fe2O3

Converting it to mole using   number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

                                           = 74g/mole

Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

                                                                           = 0.453 moles