An ore contains fe3o4 and no other iron. the iron in a 36.6-gram sample of the ore is all converted by a series of chemical reactions to fe2o3. the mass of fe2o3 is measured to be 29 g. what was the mass of fe3o4 in the sample of ore?

To solve this problem, let us first find for the molar mass of Fe2O3 and Fe3O4.

Fe = 55.85 g/mol and O = 16 g/mol

Therefore,

Fe2O3 = 159.7 g/mol

Fe3O4 = 231.55 g/mol

We are given that there are 29 g of Fe2O3, we calculate for the amount of Fe from this in moles:

mol Fe = 29 g Fe2O3 (1 / 159.7 g/mol) (2 mol Fe / 1 mol Fe2O3)

mol Fe = 0.363 mol

Converting this to Fe3O4:

mass Fe3O4 = 0.363 mol Fe (1 mol Fe3O4 / 3 mol Fe) (231.55 g/mol)

mass Fe3O4 = 28.03 g

Therefore there are 28.03g of Fe3O4 in the ore.

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