Iron produced : 1.5 kg
<h3>Further explanation</h3>Reaction
a) CaCO3→CaO + CO2
b) CO2 + C → 2CO
c) Fe2O3 + 3CO → 2Fe + 3CO2
mol CaCO₃ (MW=100,0869 g/mol) :
mol CO₂ = mol CaCO₃ =13.39(reaction a)
mol CO = 2 x mol CO₂ = 26.78(reaction b)
mol CO(reaction c) = 1.5 mol CO from reaction b = 1.5 x 26.78 = 40.17
mol Fe(reaction c) = 2/3 x mol CO =
Or we can cancel out the same compound in different sides from the sum of the three reactions to find mol Fe
a) CaCO3→CaO + CO2
b) CO2 + C → 2CO
c) Fe2O3 + 3CO → 2Fe + 3CO2
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CaCO3 + Fe2O3 + C + CO + 3CO2 ⇒ CaO + 2Fe
mol Fe = 2 x mol CaCO3 = 2 x 13.39 = 26.78
mass Fe :