Answer
The actual yield of iron in moles = 0.313 mol
The theoretical yield of iron in moles = 0.672 mol
The percent yield = 46.58%
Explanation
Given:
Moles of Fe₂O₃ that react = 0.336 mol
Mass of Fe produced = 17.5 g
Equation: Fe₂O₃ + 3C → 2Fe + 3CO
What to find:
i. The actual yield of iron in moles.
ii. The theoretical yield of iron in moles.
iii. The percent yield.
Step-by-step solution:
i. The actual yield of iron in moles.
From the given information; the actual yield of Fe in grams is 17.5 g.
So, the actual yield in moles can be determined by converting grams to mole using the mole formula.
[tex]Moles=\frac{Mass}{Molar\text{ }mass}[/tex]Using the periodic table, the molar mass of Fe = 55.845 g/mol
[tex]Moles=\frac{17.5g}{55.845g\text{/}mol}=0.313\text{ }mol[/tex]Therefore, the actual yield of iron in moles is 0.313 mol.
ii. The theoretical yield of iron in moles.
Using the mole ratio Fe₂O₃ and Fe, and the given moles of Fe₂O₃; theoretical yield of Fe in moles is calculated as follows:
[tex]\begin{gathered} 1\text{ }mol\text{ }Fe₂O₃=2\text{ }mol\text{ }Fe \\ \\ 0.336\text{ }mol\text{ }Fe₂O₃=x \\ \\ x=\frac{0.336\text{ }mol\text{ }Fe₂O₃}{1\text{ }mol\text{ }Fe₂O₃}\times2\text{ }mol\text{ }Fe \\ \\ x=0.672\text{ }mol\text{ }Fe \end{gathered}[/tex]Thus, the theoretical yield of iron in moles is 0.672 mol.
iii. The percent yield.
The percent yield is calculated using the formula:
[tex]Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\%[/tex]Putting the actual yield = 0.313 mol and the theoretical yield = 0.672 mol into the formula, we have the percent yield to be equal:
[tex]\begin{gathered} Percent\text{ }yield=\frac{0.313\text{ }mol}{0.672\text{ }mol}\times100\% \\ \\ Percent\text{ }yield.=46.58\% \end{gathered}[/tex]The percent yield is 46.58%
