Solid iron(III) oxide reacts with solid aluminum to produce solid iron and solid aluminum oxide. If 50.0 grams of iron(III) oxide were mixed with 50.0 grams of aluminum, what maximum mass of iron metal could be produced?
I got 52.5 g . Please don't just say yes without meaning it... please... I just want to make sure I did this right so I understand it.
I got 52.5 g . Please don't just say yes without meaning it... please... I just want to make sure I did this right so I understand it.
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I think the answer is actually 35g.
Start with a balanced equation.
Fe2O3 + 2Al --> 2Fe + Al2O3.
First step is to figure out how much Fe if produced by Fe2O3.
Take the initial mass, divide(multiply by 1/molar mass) by the molar mass to get the number of moles, then do a mole ratio of Fe to Fe2O3, then multiply by the molar mass of Fe.
50g * molFe2O3/159.7g * 2molFe/1molFe2O3 * 55.85g/molFe = 35g.
This means the maximum amount of Fe that can be produced by Fe2O3 is 35g.
Next, figure out how much Fe is produced by Al.
Take the initial mass of Al, divide(multiply by 1/molar mass) by the molar mass to get number of moles, then do a mol ratio of Fe to Al, then multiply by the molar mass of Fe.
50g * molAl/26.98g * 2molFe/2molAl * 55.85g/molFe = 103.5
This means that Al can produce 103.5g of Fe.
However, since the most Fe that Fe2O3 can produce is 35g, that will be the maximum amount of Fe produced.
We call Fe2O3 the limiting reagent, because it limits the amount of product that can be produced.
If you have any questions let me know and I will help =)
Start with a balanced equation.
Fe2O3 + 2Al --> 2Fe + Al2O3.
First step is to figure out how much Fe if produced by Fe2O3.
Take the initial mass, divide(multiply by 1/molar mass) by the molar mass to get the number of moles, then do a mole ratio of Fe to Fe2O3, then multiply by the molar mass of Fe.
50g * molFe2O3/159.7g * 2molFe/1molFe2O3 * 55.85g/molFe = 35g.
This means the maximum amount of Fe that can be produced by Fe2O3 is 35g.
Next, figure out how much Fe is produced by Al.
Take the initial mass of Al, divide(multiply by 1/molar mass) by the molar mass to get number of moles, then do a mol ratio of Fe to Al, then multiply by the molar mass of Fe.
50g * molAl/26.98g * 2molFe/2molAl * 55.85g/molFe = 103.5
This means that Al can produce 103.5g of Fe.
However, since the most Fe that Fe2O3 can produce is 35g, that will be the maximum amount of Fe produced.
We call Fe2O3 the limiting reagent, because it limits the amount of product that can be produced.
If you have any questions let me know and I will help =)