How many grams of Al2O3 is extracted from 250. g of FeO? Please explain

Answer 1

First start off by writing a reaction of where this would be taking place. You are given in the problem that you Al2O3 will be extracted so this gives you a clue that this may likely be on the product side of the reaction.

So if you write and equation where the Al2O3 is a product you would have a replacement type reaction where many times a metal is reacting with Oxygen to some level and then one metal is replacing the other metal.

Then you were given in the problem that you have 250g of the FeO, iron oxide and because it indicates from you can conclude that this species is more than likely on the reactant side.

So you would have to write it out in balance form, basically coming up with a balance reaction of what this would look like:

FeO + Al+3 -----> Al2O3 + Fe+2

You would have to make this balanced:

For the molecule Al2O3 there is no net charge so Al would have to take on an oxidation number of 3 or charge of +3. Oxygen rarely takes on anything other than -2 for charge and even for oxidation number. for this to be a balanced and no net charge molecule.

Fe has a 2+ charge

Al has a 3+ charge

and

O has a -2 charge

So these elements would have to come together in a way in which there would be no net charge.

Now you can write a balanced equation:

If you put the following coeffients in front of each species of the product and reactant side:

3FeO + 2Al+3 -----> Al2O3 + 3Fe+2

3 Fe on both sides

3 O on both sides

and

2 Al on both sides

From this equation you can see how all the species on the reactant and product side interact with each other.

3 moles of FeO react with 2 moles of Al+3

250g of FeO react with how many grams of Al+3

3.4797 mol of FeO how many moles of this would be equvalent for Al+3?

3 mol FeO -------> 2 moles Al+3

3.4797 FeO --------X moles Al+3

cross multiply divide and solve for x:

3x = 2(3.4797)

x= 2(3.4797)/3 = 2.3198 moles of Al+3

If trying to determine the limiting reactant you can take the moles derived from the problem of each species on the reactant side and divide it by the coeffecient in the balanced equation:

FeO: 3.4797/3 = 1.1599

Al+3: 2.3198/2 = 1.1599

They both have the same moles-given-from-problem-to-coeffecient ratio, so you might as well proceed the reaction with the grams you were given from FeO which is 250.g

now that you have a BALANCED equation you can see what was yielded or extracted from the reactants given.

If you were not given grams for the Al weighted out in the reaction and only given the grams for the FeO you would use those grams the 250g to figure out how much Al2O3 would be yielded or extracted.

You could convert the 250g of the FeO to moles.

FeO atomic weight is 55.845 + 16 = 71.845g/mol

250 g of FeO x 1mol/71.845 g = 3.4797 mol

now using the balanced equation and the coeffecients infront of each species on the reactant and product side, find out how much can be be yielded or extracted:

3FeO + 2Al+3 -----> Al2O3 + 3Fe+2

3 moles of FeO react with or react to produce 1 mole of Al2O3. There is an invisble 1 infront of the Al2O3 in the balance equation reaction:

3molesFeO ------>1molesAl2O3

3.4797molesFeO--------> XAl2O3

Now you can cross multiply and divide and solve for X:

3X= 3.4797

X= 3.4797/3 = 1.1599 moles of the Al2O3

should we stop here? No.

It wants our answer in grams.

take the atomic weight of Al2O3 and find the grams from this many moles: 1.1599

atomic weight for Al2O3 is:

Al 26.98 x 2= 53.96

O 16 x 3 = 48

101.96g/mol

1.1599 mol x 101.96g/mol

moles will cancel moles top to bottom and you will have your grams at: 118.26g but to 3 significant figures 118 g.of Al2O3.