Question

In the metallurgic industry one of the processes to get pure iron takes tree steps.
a) CaCO3→CaO + CO2


b) CO2 + C → CO


c) Fe2O3 + CO → Fe + CO2


Calculate the kilograms of iron that would be produced from 1340 g of calcium carbonate

Answer 1

Iron produced : 1.5 kg

Further explanation

Reaction

a) CaCO3→CaO + CO2

b) CO2 + C → 2CO

c) Fe2O3 + 3CO → 2Fe + 3CO2

mol CaCO₃ (MW=100,0869 g/mol) :

$\tt \dfrac{1340}{100,0869}=13.39$

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 =

$\tt \dfrac{2}{3}\times 40.17=26.78$

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

----------------------------------------------+

CaCO3 + Fe2O3 + C + CO + 3CO2 ⇒ CaO + 2Fe

mol Fe = 2 x mol CaCO3 = 2 x 13.39 = 26.78

mass Fe :

$\tt 26.78\times 55.845~g/mol=1495.53~g\approx 1.5~kg$