Answer : The mass of iron metal needed are, 87.4 grams
Solution : Given,
Mass of Al = 150 g
Mass of \(Fe_2O_3\) = 250 g
Molar mass of Al = 27 g/mole
Molar mass of \(Fe_2O_3\) = 160 g/mole
Molar mass of Fe = 56 g/mole
First we have to calculate the moles of Al and \(Fe_2O_3\).
\(\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{150g}{27g/mole}=5.55moles\)
\(\text{ Moles of }Fe_2O_3=\frac{\text{ Mass of }Fe_2O_3}{\text{ Molar mass of }Fe_2O_3}=\frac{250g}{160g/mole}=1.56moles\)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
\(Fe_2O_3+2Al\rightarrow Al_2O_3+2Fe\)
From the balanced reaction we conclude that
As, 2 mole of \(Al\) react with 1 mole of \(Fe_2O_3\)
So, 1.56 moles of \(Al\) react with \(\frac{1.56}{2}=0.78\) moles of \(Fe_2O_3\)
From this we conclude that, \(Fe_2O_3\) is an excess reagent because the given moles are greater than the required moles and \(Al\) is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of \(Fe\)
From the reaction, we conclude that
As, 2 mole of \(Al\) react to give 2 mole of \(Fe\)
So, 1.56 mole of \(Al\) react to give 1.56 mole of \(Fe\)
Now we have to calculate the mass of \(Fe\)
\(\text{ Mass of }Fe=\text{ Moles of }Fe\times \text{ Molar mass of }Fe\)
\(\text{ Mass of }Fe=(1.56moles)\times (56g/mole)=87.4g\)
Thus, the mass of iron metal needed are, 87.4 grams
