Limiting Reagent Issue

[MarinaD]

Hello, so i understand how to calculate the limiting reagent given a certain reaction. For example,
2 Al(s) + 3 CuCl2(aq) → 2 AlCl3(aq) + 3 Cu(s)

I was given the grams of aluminum (2g) and the volume of the solution ( 500 mL CuCl2)

However, when doing the mole ratio calculations, how do i know which product to use? Like after converting my Al to mols, would my mol ratio be with Al and AlCl3 or Cu? Or can I use both?

Lastly, are there any shortcuts for this for the MCAT? I have issues dividing decimal numbers.

Thanks in advance.

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[Swagster]

The trick it to compare the given ratio of Al:CuCl2 to the needed ratio of 2:3. If the ratio is bigger than 2:3, you have excess Al and will run out of CuCl2 first. If the ratio is less than 2:3, then you have excess CuCl2 and will run out of Al first.

You have 2/27 moles Al and (M x 0.5) moles CuCl2. Calculate that ratio and compare it to 2:3 and you're golden.

 

[rdyotz]

It doesn't matter which you use, so keep it standard, IE: Do not use compare the number of moles of Al used to create moles of AlCl3 to the number of moles used to make x moles of Cu, as the ratios are different. This actually seems more confusing the way I wrote it, so what I mean to say is choose 1 product to compare both of your initial amounts to.

So you calculate the number of moles Al you start with and the number of moles of CuCl2 you start with. Then using each of those numbers, you calculate the number of moles of AlCl3 (or you can use Cu instead) you end up with.

2g Al / 26.98(g/mol) Al = 0.07moles Al
0.5L CuCl2 * molarity CuCl2 = x moles CuCl2

0.07mol Al * (2 mol AlCl3 / 2 mol Al) = 0.07 mol Al
x mol CuCl2 * (2 mol AlCl3 / 3 mol CuCl2) = 2/3 * x mol CuCl2

As long as you use the same product for both calculations you will find the limiting reagent. If you are asked to determine exactly how much of a certain product is attained, then obviously use that product in the calculations. But, the limiting reagent is simply the reactant that results in a lower product.

As for the math, simplify things. Moles of Al (2/26.98) can just be 2/27 (mental math, you can multiple by 4/4 to get 8/108, which means it would be slightly less than 0.08 (8/100), so say 0.075. Which is really close to the actual number of 0.074, roughly 1% off. This is close enough. You can always do long division, but that shouldn't be necessary.

I feel like I rambled and not sure how well I answered the actual question anymore, so let me know if there are any questions.

 

[MarinaD]

It doesn't matter which you use, so keep it standard, IE: Do not use compare the number of moles of Al used to create moles of AlCl3 to the number of moles used to make x moles of Cu, as the ratios are different. This actually seems more confusing the way I wrote it, so what I mean to say is choose 1 product to compare both of your initial amounts to.

So you calculate the number of moles Al you start with and the number of moles of CuCl2 you start with. Then using each of those numbers, you calculate the number of moles of AlCl3 (or you can use Cu instead) you end up with.

2g Al / 26.98(g/mol) Al = 0.07moles Al
0.5L CuCl2 * molarity CuCl2 = x moles CuCl2

0.07mol Al * (2 mol AlCl3 / 2 mol Al) = 0.07 mol Al
x mol CuCl2 * (2 mol AlCl3 / 3 mol CuCl2) = 2/3 * x mol CuCl2

As long as you use the same product for both calculations you will find the limiting reagent. If you are asked to determine exactly how much of a certain product is attained, then obviously use that product in the calculations. But, the limiting reagent is simply the reactant that results in a lower product.

As for the math, simplify things. Moles of Al (2/26.98) can just be 2/27 (mental math, you can multiple by 4/4 to get 8/108, which means it would be slightly less than 0.08 (8/100), so say 0.075. Which is really close to the actual number of 0.074, roughly 1% off. This is close enough. You can always do long division, but that shouldn't be necessary.

I feel like I rambled and not sure how well I answered the actual question anymore, so let me know if there are any questions.

That made sense! Thank you, the math tip really helped. Will multiplying the numerator and denominator by the same value to simplify my answer always work?

 

[MarinaD]

The trick it to compare the given ratio of Al:CuCl2 to the needed ratio of 2:3. If the ratio is bigger than 2:3, you have excess Al and will run out of CuCl2 first. If the ratio is less than 2:3, then you have excess CuCl2 and will run out of Al first.

You have 2/27 moles Al and (M x 0.5) moles CuCl2. Calculate that ratio and compare it to 2:3 and you're golden.

Thank you!

 

[rdyotz]

That made sense! Thank you, the math tip really helped. Will multiplying the numerator and denominator by the same value to simplify my answer always work?

Multiplying by the same value, ie: 4/4, is just the same as multiplying by 1 (1/1), so that will always work. It is more just practicing and finding what will work best. If you have 19/34, then multiplying 4/4 would not be ideal, but 3/3 would give you 57/102 which would be a good number. Everyone has a different set of numbers they feel comfortably with, I typically try and get the final number (57/102) close to 100 in the denominator for easy calculations. Some people do fine getting near the decimal with 21/34 (this isn't hard either, roughly 2/3). Until you get the hang of it, it may take a couple of tries/calculations.

 

[MarinaD]

Multiplying by the same value, ie: 4/4, is just the same as multiplying by 1 (1/1), so that will always work. It is more just practicing and finding what will work best. If you have 19/34, then multiplying 4/4 would not be ideal, but 3/3 would give you 57/102 which would be a good number. Everyone has a different set of numbers they feel comfortably with, I typically try and get the final number (57/102) close to 100 in the denominator for easy calculations. Some people do fine getting near the decimal with 21/34 (this isn't hard either, roughly 2/3). Until you get the hang of it, it may take a couple of tries/calculations.

this is awesome, thanks