Answer to Question #306224 in General Chemistry for Ali

Question

Iron produces from fe2o3 by reaction with carbon. The reaction products are

Fe, FeO and CO. If we mix 100 Kg of Fe2o3 and 24 Kg of C and produced 12 Kg of FeO, determine:

a) Limiting reaction and excess reaction.

b) % Excess reaction.

c) % Conversion of Fe2o3 and C.

d) The degree of completion of reaction.

e) Mass of Fe in Kg.

Fe2o3+ 2C / Fe+FeO+2CO

Mw;

Fe = 56

C=12.

O = 16

Answer 1

Fe₂O₃ + 2C → Fe + FeO + 2CO

a)      Limiting reagent:

Molar mass Fe₂O₃: 159.69

Moles Fe₂O₃: 100 / 159.69 = 0.626 kmol

Molar mass C: 12.01

Moles Fe₂O₃: 24 / 12.01 = 2.00 kmol

The mole ratio according to the equation: 1 (Fe₂O₃) to 2 (C); meaning 2.00 mole C would require 1.00 mole (Fe₂O₃). As we have less (0.626), we conclude that Fe₂O₃ is limiting reagent.

  

b)     % Excess reagent:

C is in excess; thus, it is an excess reagent.

 

c)      % Conversion of Fe₂O₃ and C:

100% of Fe₂O₃ is used up, because it is a limiting reagent;

0.626 kmol x 2 = 1,252 kmol (C) reacts

1.252 / 2.00 = 0,626 or 62,6% - converted C.

 

d)     The degree of completion of reaction:

Molar mass FeO: 71.84

Moles FeO: 12 / 71.84 = 0.167 kmol

Theoretically we can obtain: 0.626 kmol (same as Fe₂O₃)

0.167 / 0.626 = 0.27 or 27% - the degree of completion.

 

e)     Mass of Fe in kg:

Moles of Fe: 0.167 kmol

Molar mass Fe: 55.85

Mass Fe: 55.85 x 0.167 = 9.33 kg.