Question

Mining companies use this reaction to obtain iron from iron ore:

Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g)

The reaction of 187 g
of Fe2O3
with 82.4 g
of CO
produces 70.5 g
of Fe
.
Determine the percent yield of solid iron.

Answer 1

Answer:Answer:

The theoretical yield of solid iron is 110.05 grams

Explanation: 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