Question

5.0 g of iron is reacted with 5.0g of water according to the chemical equation shown below. 3 Fe(s) + 4 H2O(l) → Fe3O4(s) + 4 H2(g)

a) Calculate the amount of Fe3O4 produced, in grams.

b) What is the mass of excess reagent, in grams?

Answer 1

Fe (s) +   4H2O (l) --> Fe3O4 (s) + 4H2 (g)

Iron moles = mass / Molar mass of Iron

              = 5 g / 55.845 g/mol

        = 0.08954

H2O moles = 5 g / ( 18.015 g/mol)

           = 0.27755

for 1 Fe 4 H2O is needed

hence for 0.08954 moles Fe , H2O moles needed = 4 x 0.08954 = 0.358

but we had only 0.2775 moles H2O

a) hence H2O is limiting reagent

Fe3O4 moles produced = ( 1/4) H2O moles = ( 1/4) 0.2775 = 0.0694

Fe3O4 mass produced = moles xmolar mass of Fe3O4

               = 0.0694 mol x 231.533 g/mol

          = 16 g

b) Fe moles reacted = ( 1/4) H2O moles = ( 1/4) ( 0.2775) = 0.0694

Fe moles left unreacted = initial Fe moles - Fe moles reacted

                     = 0.08954 - 0.0694 = 0.02014

Fe mass excess = Fe moles excess x molar mass of Fe

                = 0.02014 mol x 55.845 g/mol

          = 1.125 g