An iron nail rusts when exposed to oxygen. According to the following reaction, how many grams of iron(III) oxide will be formed upon the complete reaction of 31.1 grams of oxygen gas with excess iron?

iron(s) + oxygen(g) → iron(III) oxide(s)

moles iron(III) oxide

103.8 g of , 0.65 moles of

Explanation:

To calculate the moles :

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

According to stoichiometry :

3 moles of produce =  2 moles of

Thus 0.975 moles of will produce= of

Mass of

Thus 103.8 g of will be produced.

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answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole