Answer:
The theoretical yield of solid iron is 110.05 grams
Explanation:
<u>Step 1</u>: 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
<u>Step 2</u>: 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
<u>Step 3:</u> 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)
<u>Step 4</u>: 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
<u>Step 5:</u> 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
