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### Topic: Problem with Limiting Reactants  (Read 1360 times)

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#### B9766

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##### Problem with Limiting Reactants
« on: October 11, 2018, 09:57:31 PM »
I have a "limiting reactants" textbook problem for which I can't get the same answer as the book.
Aluminum and bromine vigorously react according to the equation - $2Al(s)\ +\ 3Br_{2}(l)\ \rightarrow \ 2AlBr_{3}(s)$

(a) If 5.00 g of aluminum and 22.2 g of bromine react, what mass of AlBr3 is produced? - (Book's answer is 24.7 g)
(b) What mass of the excess reactant remains at the end of the reaction? - (Book's answer is 2.50 g)

My solution:
$5.00g\ Al\ *\ \dfrac{1\ mol \ Al}{26.98g}*\dfrac{2\ mol\ AlBr_{3}}{2\ mol\ Al}*\dfrac{266.68g \ AlBr_{3}}{1\ mol\ AlBr_{3}}=49.2g\ AlBr_{3}$

$22.2g\ Br\ *\ \dfrac{1\ mol \ Br}{79.90g}*\dfrac{2\ mol\ AlBr_{3}}{3\ mol\ Br}*\dfrac{266.68g \ AlBr_{3}}{1\ mol\ AlBr_{3}}=49.2g\ AlBr_{3}$

They came out exactly the same but my answer is about 2x the one given in the book. And since they both come out to the same answer I can't see where one would be limiting.

If I plug the book's answer (24.7g AlBr3) into the equation to solve for Al, I get:
$24.7g \ AlBr_{3} * \dfrac{1\ mol\ AlBr_{3}}{266.68g} *\ \dfrac{2\ mol\ Al}{2\ mol\ AlBr_{3}}\ *\ \dfrac{29.68g\ Al}{1\ mol\ Al}\ =\ 2.75g\ Al$

Definitely not the 5.00g Al I started out with. So, whatever I'm doing wrong is consistently wrong.
Any help would be appreciated.

#### XeLa.

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##### Re: Problem with Limiting Reactants
« Reply #1 on: October 12, 2018, 12:49:59 AM »
Hi B9766!

The main issue here comes down to identifying what type of question this is. You state it explicitly yourself:

I have a "limiting reactants" textbook problem

Have you come across such problems before? If so, you should recognise that one of the reactants is in excess whilst the other is a limit - that is, it reacts completely with the excess reactant to give the product, leaving some of the excess reactant left over.

Can you identify which reactant is the limit?

Also, as a side note, make sure that you are using the molar mass of Br2 not Br in your second line of equations! The bromine that the textbook is talking about is Br2.

#### Borek

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##### Re: Problem with Limiting Reactants
« Reply #2 on: October 12, 2018, 02:46:58 AM »
To add to what XeLa wrote - calculate, how much Br2 would react with 5 g of Al. Do you have that much?

Calculate, how much Al would react with 22.2 g of Br2. Do you have that much?
Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### mjc123

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##### Re: Problem with Limiting Reactants
« Reply #3 on: October 12, 2018, 04:54:52 AM »

22.2g Br ∗ 1 mol Br/79.90g∗2 mol AlBr3/3 mol Br∗266.68g AlBr3/1 mol AlBr3=49.2g AlBr3

That's your problem. It should be 22.2g Br2, not Br.

#### B9766

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##### Re: Problem with Limiting Reactants
« Reply #4 on: October 12, 2018, 10:16:29 AM »
Ah! Thank all, especially XeLa. Yes, I've already done several "limiting reactant" problems and understand how they are done - except when one of the reactants is diatomic, obviously. I should have caught that.