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1 November, 18:25

Combining 0.320 mol Fe2O3 with excess carbon produced 10.8 g Fe. Fe 2 O 3 + 3 C ⟶ 2 Fe + 3 CO a. What is the actual yield of iron in moles? b. What is the theoretical yield of iron in moles? c. What is the percent yield?

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  1. 1 November, 18:50
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    a. 0.193 moles of Fe, actual yield

    b. 0.640 moles of Fe, the theoretical yield

    c. 30 % yield

    Explanation:

    The reaction is this: Fe₂O₃ + 3C → 2Fe + 3CO

    First of all, we determine the moles of produced Fe.

    10.8 g / 55.85 g/mol = 0.193 moles → Actual yield of iron in moles

    We have 0.320 moles, so we can calculate the theoretical yield by a rule of three.

    1 mol of Fe₂O₃ can produce 2 moles of Fe

    0.320 moles of oxide, may produce (0.320. 2) / 1 = 0.640 moles

    As we determined the theoretical yield and the actual yield, we can know the percent yield → (Produced yield / Theoretical Yield). 100 = % yield

    (0.193 mol / 0.640 mol). 100 = 30%
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