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stellarik [79]
3 years ago
5

Fe2O3 + 2Al -> Al2O3 + 2Fe

Chemistry
2 answers:
pashok25 [27]3 years ago
7 0
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 
svet-max [94.6K]3 years ago
5 0

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

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