1) In a blast furance, a mixture of iron (iii) oxide ( Fe2O3 ) and carbon is hea

Question

1) In a blast furance, a mixture of iron (iii) oxide ( Fe2O3 ) and carbon is heated to form iron metal and carbon dioxide. A) write and balance a chemical equation for this reaction. B) how many kilograms of iron can be produced by the reaction of 7.00 kg of fe2o3 with excess carbon? C) how many kilograms of iron can be produced by the reaction of 1.85 kg of carbon with excess Fe2O3 ? D) how many kilograms of iron can be produced by the reaction of 7.00 kg of Fe2O3 with 1.85 kg of carbon? E) After reacting 7.00 kg of Fe2 O3 and 1.85 kg of carbon, 4.80 kg of Fe were recovered. What is the precent yield of this reaction? (( SHOW WORKS PLEASE ! ))

Explanation / Answer

A) 2Fe2O3 + 3C ---> 4Fe + 3CO2

M.wt of Fe2O3 = 156.69 g/mol , Fe = 55.85 g, C = 12 g/mol , CO2 = 44.01 g/mol

B) 7.0 kg of Fe2O3

7.0 kg of Fe2O3 = 7000 g / 156.69 g/mol = 44.67 mol ( 2*22.33 mol)

Therefore 4*22.33 mol Fe would be formed = (4*22.33) mol *55.85 g/mol = 4.99 kg of Fe

C) 1.85 kg of Carbon

1.85 kg of Carbon = 1850 g / 12 g/mol = 154.17 mol ( 3*51.39 mol)

Therefore 4*51.39 mol of Fe would be formed = ( 4*51.39 mol )* 55.85 g/mol = 11.48 kg of Fe

D) Reaction of 7.00 kg of Fe2O3 with 1.85 kg of carbon

7.0 kg of Fe2O3 = 7000 g / 156.69 g/mol = 44.67 mol ( 2*22.33 mol)

1.85 kg of Carbon = 1850 g / 12 g/mol = 154.17 mol ( 3*51.39 mol)

Fe2O3 is limitting reagent and Carbon is excess, so *22.33 mol Fe would be formed

(4*22.33) mol *55.85 g/mol = 4.99 kg of Fe

E) If 4.80 kg of Fe were recovered, then the % yield is (4.80 kg /4.99 kg)*100 = 96.19

[ If you get 4.99 kg , then it is 100 % yield]

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