In the reaction given below, 23.5 grams of iron(III) oxide (Fe2O3) are reacted with excess carbon. What is the theoretical yield of iron metal? 2 Fe2O3 + 3 C --> 4 Fe + 3 CO2

If 14.5 grams of iron are recovered when the reaction is complete, what is the percent yield?

**Both answer should be written as XX. X

Theoretical yield = 16.76 g

Percent yield = 87%

Explanation:

Given data:

Mass of iron oxide = 23.5 g

Theoretical yield of iron = ?

Actual yield of iron = 14.5 g

Percent yield of iron = ?

Solution:

Chemical equation:

2Fe₂O₃ + 3C     →  4Fe + 3CO₂

Number of moles of iron oxide:

Number of moles = mass/molar mass

Number of moles = 23.5 g/ 159.69 g/mol

Number of moles = 0.15 mol

Now we will compare the mole of iron oxide with iron metal.

                      Fe₂O₃         :            Fe

                         2              :              4

                      0.15            :          4/2×0.15 = 0.3

Mass of iron:

Mass = number of moles × molar mass

Mass = 0.3 mol × 55.85 g/mol

Mass = 16.76 g

Percent yield:

percent yield = (actual yield / theoretical yield)× 100

Percent yield = (14.5 g / 16.76 g) × 100

Percent yield = 0.87 × 100

Percent yield = 87%

the answer is b & c i believe : )

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