Question

If 0.450 moles of iron III oxide (Fe2O3) are allowed to react with an excess of aluminum (Al) and 43.6 grams of iron (Fe) is produced, what is the percent yield of iron?
2Al + Fe2O3 2Fe + Al2O3

Answer 1

21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe 
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe 
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced. 

Answer 2

Answer : The percent yield of iron is, 86.5 %

Explanation : Given,

Moles of iron(III)oxide = 0.450 mole

Mass of iron = Actual yield of Fe = 43.6 g

First we have to calculate the moles of iron.

The balanced chemical reaction is,

$2Al+Fe_2O_3\rightarrow 2Fe+Al_2O_3$

From the balanced reaction we conclude that

As, 1 mole of $Fe_2O_3$ react to give 2 mole of $Fe$

So, 0.450 mole of $Fe_2O_3$ react to give $\frac{2}{1}\times 0.450=0.9$ moles of $Fe$

Now we have to calculate the mass of Fe.

$\text{Mass of }Fe=\text{Moles of }Fe\times \text{Molar mass of }Fe$

$\text{Mass of }Fe=(0.9mole)\times (56g/mole)=50.4g$

Therefore, the mass iron produces, 50.4 g

Now we have to calculate the percent yield of Fe.

$\%\text{ yield of }Fe=\frac{\text{Actual yield of }Fe}{\text{Theoretical yield of }Fe}\times 100=\frac{43.6g}{50.4g}\times 100=86.5\%$

Therefore, the percent yield of Fe is, 86.5 %