Question

PLZ HELP!!!!!!!!!!!!!

Answer 1

117.22 g are needed to react with an excess of Fe2O3 to produce 156.2 g of Fe.

Explanation:

                  Moles of Fe = Mass of Fe in grams / Atomic weight of Fe

                                        = 156.2 / 55.847

                  Moles of Fe  = 2.79.

The ratio between CO and Fe id 3 : 2.

                  Moles CO needed  = 2.79 * (3 / 2)

                                                   = 4.185.

 To calculate Atomic weight of CO,

                  Atomic weight of carbon = 12.011

                  Atomic weight of oxygen= 15.9994

        Atomic weight of CO = 12.011 + 15.9994 = 28.01 g / mol.

                 Mass of CO   = 4.185 * 28.01 = 117.22 g.