Mining companies extract iron from iron ore according to the following balanced equation: Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g) In a reaction mixture containing 175 g Fe2O3 and 73.0 g CO, CO is the limiting reactant. Calculate the mass of the reactant in excess (which is Fe2O3) that remains after the reaction has gone to completion. Express the mass with the appropriate units.

There will remain 36.7 grams of Fe2O3

Explanation:

Step 1: Data given

Mass of Fe2O3 = 175 grams

Mass of CO = 73.0 grams

CO is the limiting reactant

Molar mass of CO = 28 g/mol

Molar mass of Fe2O3 = 159.69 g/mol

Step 2: The balanced equation

Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g)

Step 3: Calculate moles of CO

Moles CO = mass CO / molar mass CO

Moles CO = 73.0 grams / 28.0 g/mol

Moles CO = 2.61 moles

Step 4: Calculate moles of Fe2O3

Moles Fe2O3 = 175.0 grams / 159.69 g/mol

moles Fe2O3 = 1.10 moles

Step 5: Calculate moles reacted

Co is the limiting reactant. It will completely be consumed (2.61 moles)

Fe2O3 is in excess. There will be 2.61/3 = 0.87 moles of Fe2O3 consumed

There will remain 1.10 - 0.87 = 0.23 moles of Fe2O3

Step 6: Calculate the mass of Fe2O3 that remains

Mass Fe2O3 = moles * molar mass

Mass Fe2O3 = 0.23 moles * 159.69 g/mol

Mass of Fe2O3 = 36.7 grams

There will remain 36.7 grams of Fe2O3