Question

identify the limiting reactant in the reaction of iron and oxygen to form Fe2O3 if 44.1 g of Fe and 17.1 g if O2 are combined. Determine the amount ( in grams) of excess reactant that remains after the reaction is complete

Answer 1

4 Fe    + 3 O2      ----------------> 2 Fe2O3

223.38 g          96 g                                  319.38 g

44.1 g              17.1 g                                 ??

here

223.38 g Fe    ---------------> 96 g O2

44.1 g Fe        --------------->   ??

mass of O2 needed = 44.1 x 96 / 223.38 = 18.95 g O2 .

but we have only 17.1 g O2 . so here

limiting reagent is O2 .

excess reagent is Fe.

excess remain = 44.1 - (17.1 x 223.38 / 96) = 4.31 g

excess reactant that remains = 4.31 g