Answer:
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
Iron(III) oxide reacts with carbon monoxide to produce iron along with carbon dioxide as the by-product. From the reaction, the theoretical yield of iron is 110.05 grams.
A theoretical yield is the amount of the product that is produced and calculated by the stoichiometry coefficient. It differs from the actual yield as it is actually produced.
The chemically balanced reaction:
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
From above, 1 mole of iron (III) oxide = 3 moles of carbon dioxide and 1 mole of iron.
Moles of Ferric (III) oxide: 177 ÷ 159.69 = 1.1084 moles
Moles of carbon monoxide: 82.8 ÷ 28.01 = 2.9561 moles
Here, Ferric (III) oxide is the limiting reagent and carbon monoxide is the excess reagent. Moles of Iron are calculated as 1.9707 moles.
Mass of iron:
mass = moles × Molar mass
1.9707 × 55.845
= 110.05 gms
Therefore, 110.05 grams is the theoretical yield of iron.
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