Question

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:

Answer 1

Answer:

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

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 $Fe_2O_3$ = 0.322 moles

According to the given reaction:

$Fe_2O_3+3C\rightarrow 2Fe+3CO$

1 mole of $Fe_2O_3$ on reaction produces 2 moles of Fe

0.322 mole of $Fe_2O_3$ 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 %