Question

Fe2O3 + 2Al -> Al2O3 + 2Fe

Calculate the mass of iron metal (in grams) that can be prepared from 150 grams of aluminum and 250 grams of iron(III) oxide.

Help please!

Answer 1

Fe2O3 + 2Al ---> Al2O3 + 2Fe 
Mole ratio Fe2O3 : Al = 1:2 
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles 
No. of moles of Al = 150/27 = 5.555555555 moles. 
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react. 

So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared. 
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe 

Answer 2

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