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

ID: 1004390 • Letter: 1

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?

A. 2Fe2O3 + 3C -------> 2Fe + 3CO2

B. from balanced equation

2 moles of Fe2O3 react with C to form 4 moles of Fe

2*159.7g of Fe2O3 react with C to form 4*55.84g of Fe

7Kg of Fe2O3 react with C to form = 4*55.84*7/2*159.7 = 4.9Kg of Fe

C. 2Fe2O3 + 3C -------> 2Fe + 3CO2

3 moles of carbon react with Fe2O3 to form 4 moles of Fe

3*12g of carbon react with Fe2O3 to form 4*55.84g of Fe

1.85kg of carbon react with Fe2O3 to form = 4*55.84*1.85/3*12 = 11.48kg of Fe

D. 2Fe2O3 + 3C -------> 2Fe + 3CO2

no of moles of Fe2O3 = W/G.M.Wt = 7000/159.7 = 43.84 moles

no of moles of C = W/G.A.Wt

= 1850/12 = 154.16 moles of C

carbon is exess reagent, Fe2O3 is limiting reagent.

2 moles of Fe2O3 react with carbon to form 4 moles of Fe

2*159.7 g of Fe2O3 react with carbon to form 4*55.84g of Fe

7 kg of Fe2O3 react with carbon to form = 4*55.84*7/2*159.7 = 4.9kg of Fe

E. percntage yield of Fe = actual yield*100/theoritical yield

= 4.8*100/4.9 = 97.9%

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