Question

if you begin with 3.46 g of Fe, what mass in grams of O2 is required for complete reaction?

what is the theoretical yield of FeO3 in grams?

if 4.83 g of Fe2O3 are actually obtains, wht is the percent yield?

Answer 1

2Fe + 3O2 .......................> 2FeO3

no .of moles of Fe = wt/mol.wt = 3.46/55.84 = 0.062

As per the above equation

2moles of Fe is required 3moles of O2

0.062 moles of Fe required .......................?

    = 0.062*3/2 = 0.093 moles of O2

wt of O2 = mol.wt of O2 * no.of moles of O2 = 0.093*32 = 2.98 grams or 3.0grams

............................................................................................................................................

2 moles of Fe produce .................... 2moles of FeO3

0.062 moles of Fe ..........................................?

= 0.062 moles of FeO

so wt of FeO = no.of moles * mol.wt = 0.062*103.84 = 6.44 gm.

theoretical yield = 6.44 gms.

..........................................................................................................................

percent of yield = (practical yield/theoretical yield )*100

                        = (4.83/6.44)*100 = 75%