117.22 g are needed to react with an excess of Fe2O3 to produce 156.2 g of Fe.
Step-by-step 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.