Mining companies use this reaction to obtain iron from iron ore: fe2o3(s)+3co(g) → 2fe(s)+3co2(g) the reaction of 177 g of fe2o3 with 82.8 g of co produces 72.7 g of fe. calculate the theoretical yield of solid iron.

110. g fe

The theoretical yield of solid iron is 110.05 grams

Explanation:

Step 1: The balanced equation

Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g)

This means for 1 mole of Fe2O3 consumed, there is 3 moles of CO needed to produce 2 moles of Fe and 3 moles of CO2

Step 2: given data

Mass of Fe2O3 = 177g

MM of Fe2O3 = 159.69g/mole

Mass of CO = 82.8g

MM of CO = 28.01

Mass of CO2 = 72.7g

MM of CO2 = 44.01 g/mole

Mass of Fe = TO BE DETERMINED

MM of Fe = 55.845 g/mole

Step 3: Calculating moles

Moles = mass / Molar mass

moles Fe2O3 = 177/159.69 = 1.1084 moles

moles CO = 82.8/28.01 = 2.9561 moles

⇒ CO is the limiting reactant ; Fe2O3 is the reactant in excess (there will remain 0.123 moles)

Step 4: Calculating moles of Fe

There will be produced 2 moles of Fe if there is consumed 3 moles of CO

If there is consumed 2.9561 moles of Fe, there is produced 1.9707 moles of Fe

Step 5: Calculating mass of Fe

mass = moles * Molar mass

mass of Fe = 1.9707 moles * 55.845 = 110.05 grams

The theoretical yield of solid iron is 110.05 grams

diaria maybe go see a doctor about that