Destroyer GCHEM # 4

[td4azklz]

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The question says that 76 g of Cr2O3 reacts with 14 g of Al to produce Al2O3 and Cr metal. How many grams of Al2O3 could be produced theoretically?

The question seems so simple, and the first time i did it, my answer matched the destroyer answer (26g), but after i looked at it more, it doesnt make sense. They say that the limiting reagent is Al, however according to my stoicheometric calculations, it's supposed to be Cr2O3. What am I missing? Thanks

 

[nixon13]

This is a limiting reagent question. Start out with each of your reactants and see how much of the desired product you can get. Then the one which produces less is the limiting reagent and thats the max that can be produced of the desired product.

 

[nixon13]

Your balanced equation should look like this:

Cr2O3 + 2Al ---> Al2O3 + 2Cr

 

[td4azklz]

Your balanced equation should look like this:

Cr2O3 + 2Al ---> Al2O3 + 2Cr

I did that, and that was my equation. According to my calculations, 14g of Al need 39.4 grams of Cr2O3 and 76grams of Cr2O3 need 39 grams of Al to react with. So what does that tell us? Thanks

 

[td4azklz]

I did that, and that was my equation. According to my calculations, 14g of Al need 39.4 grams of Cr2O3 and 76grams of Cr2O3 need 39 grams of Al to react with. So what does that tell us? Thanks

I mean, 27 grams of Al are needed to react with 76 gr of CR2O3....sorry

 

[td4azklz]

OOOOh, ok...i see it now. Never mind. Thanks anyways😀

 

[Kaka89]

I was solving #44 and ran into the same problem - according to my calculations I got the right answer but thought Cr2O3 was the limiting reagent even after I balanced the equation.
Also according to Destroyer's solutions it says that you need 1 mole of Al as opposed to 0.5 moles of Cr2O3, then how is Al the limiting reagent? :/
I'm obviously missing something here.. so I'd really appreciate some help 🙂

Thanks a lot 🙂

 

[invictusx]

I was solving #44 and ran into the same problem - according to my calculations I got the right answer but thought Cr2O3 was the limiting reagent even after I balanced the equation.
Also according to Destroyer's solutions it says that you need 1 mole of Al as opposed to 0.5 moles of Cr2O3, then how is Al the limiting reagent? :/
I'm obviously missing something here.. so I'd really appreciate some help 🙂

Thanks a lot 🙂

This is what I think it is.

So, you have 0.5 moles of Cr2O3, that's the total amount of Cr2O3 you have. That amount reacted with 14 g of Al (0.5 moles). However, stoichiometrically, it's saying that you need 1 mole of Al in order for the whole 0.5 moles of Cr2O3 to react but you only have 0.5 moles of Al, so Al is the limiting reagent.

I had the same problem as well. Not sure if it's entirely correct but I hope this can take you somehwere

 

[ChrisM07]

Sorry, gotta resurrect this thread. I also realize that this is a simple problem, except something just isn't clicking. I get the part about finding out how many moles of each there are in the reaction. But Destroyer (#49, 2011) says that the next step is to compare mole ratios to determine the limiting reagent. My question is, how do you know which way to set up the ratio? They do it like this:

0.5 moles Cr2O3 x (2 moles Al/1 mole Cr2O3) = 1 mole Al needed.

But why isn't the ratio set up the other way?:

0.5 moles Al x (1 mole Cr2O3/2 moles Al) = 0.25 moles Cr2O3 needed

From here I know that we set it up like any regular stoic. problem and solve for grams of Al2O3.

Thanks a ton!

 

[csulapredental]

This is a limiting reagent question. Start out with each of your reactants and see how much of the desired product you can get. Then the one which produces less is the limiting reagent and thats the max that can be produced of the desired product.

how do u know for problems like this u need to find limiting reagent. i just would of taken my Al grams and converted to mole and then converted to grams of al2o3 what tells u to first do the limiting reagent stuff

 

[Lnguyen7]

how do u know for problems like this u need to find limiting reagent. i just would of taken my Al grams and converted to mole and then converted to grams of al2o3 what tells u to first do the limiting reagent stuff

76 g of Cr2O3 reacts with 14 g of Al
You are given two masses. It can be the same or different masses/moles.