Answered

Mining companies use this reaction to obtain iron from iron ore:
Fe2O3(s) + 3 CO(g) – 2 Fe(s) + 3 CO2(8)
The reaction of 167 g Fe2O3 with 85.8 g CO produces 72.3 g Fe. Determine the limiting reactant, theoretical yield, and
percent yield.

Answer :

Oseni

The limiting reactant, theoretical yield, and percent yield would be CO, 114.035 g, and 63.4% respectively.

From the balanced equation of the reaction:

[tex]Fe_2O_3_{(s)} + 3 CO_{(g)} ---> 2 Fe_{(s)} + 3 CO_2_{(g)}[/tex]

1 mole of Fe2O3 requires 3 moles of CO

mole of Fe2O3 = mass/molar mass

                        = 167/159.69

                         = 1.046 moles

mole of CO = 85.8/28.01

                      = 3.063 moles

1 mole Fe2O3 = 3 moles CO

1.046 moles Fe2O3 = 3 x 1.046

                                = 3.138 moles of CO

  • 3.138 moles of CO is required but only 3.063 is available. Thus, the limiting reactant is CO.

Theoretically,

1 mole Fe2O3 = 3 moles CO = 2 moles Fe

With only 3.063 moles of CO available, moles of Fe produced would be

  = 3.063x 2/3

         = 2.042 moles of Fe

Mass of 2.042 Fe = 2.042 x 55.845

                  = 114.035 g

  • Hence, theoretically, 114.035 g of Fe should be produced.

Percent yield = actual yield/theoretical yield x 100

                      = 72.3/114.035 x 100

                          = 63.4%

More on limiting reactants can be found here: https://brainly.com/question/14225536

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