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

Iron produced : 1.5 kg

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

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

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

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

mass Fe :

the answer for sodium is 0.186

the answer for rubidium is 0.251