Combining 0.322 mol fe2o3 with excess carbon produced 10.0 g fe. fe2o3+3c⟶2fe+3co what is the actual yield of iron in moles? actual yield: mol what is the theoretical yield of iron in moles? theoretical yield: mol what is the percent yield? percent yield:

Actual yield = 0.1791 moles

Theoretical yield in moles = 0.644 moles

Percent yield = 27.8 %

Explanation:

The formula for the calculation of moles is shown below:

Given: For Fe

Given mass = 10.0 g

Molar mass of Fe = 55.845 g/mol

Moles of Fe = 10.0 g / 55.845 g/mol = 0.1791 moles

So, Actual yield = 0.1791 moles

Given, Moles of = 0.322 moles

According to the given reaction:

1 mole of on reaction produces 2 moles of Fe

0.322 mole of on reaction produces 2*0.322 moles of Fe

Actual moles of Fe produced = 0.644 moles

Theoretical yield in moles = 0.644 moles

Or, Mass of iron = Moles × Molar mass = 0.644 × 55.845 g = 35.96 g

Theoretical yield = 35.96 g

Given experimental yield = 10.0 g

% yield = (Experimental yield / Theoretical yield) × 100 = (10.0/35.96) × 100 = 27.8 %

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