Question

When 50.0 g iron(III) oxide reacts with carbon monoxide, 32.5 g iron is produced. What is the percent yield of the reaction?

Answer 1

Answer:

92.9%

Explanation:

You have been given the actual yield of the reaction. First, you need to find the theoretical yield of the reaction. To do this, you need to (1) convert grams Fe₂O₃ to moles Fe₂O₃ (via molar mass from periodic table values), then (2) convert moles Fe₂O₃ to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass).

Once you have found the theoretical yield, you need to use the percent yield equation to calculate the final answer. This number should have 3 sig figs to match the given values.

(Step 1)

Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)

Molar Mass (Fe₂O₃): 159.684 g/mol

1 Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)

Molar Mass (Fe): 55.845 g/mol

50.0 g Fe₂O₃           1 mole              2 moles Fe           55.845 g
--------------------  x  ------------------  x  ---------------------  x  ----------------  = 35.0 g Fe
                               159.684 g         1 mole Fe₂O₃          1 mole

(Step 2)

                                 Actual Yield
Percent Yield  =  ---------------------------  x  100%
                             Theoretical Yield

                                32.5 g Fe
Percent Yield  =  ----------------------  x  100%  =  92.9%
                                35.0 g Fe