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Author Topic: lmiting reagent calculation for maleic anhydride  (Read 16818 times)

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phillyj

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lmiting reagent calculation for maleic anhydride
« on: February 06, 2008, 12:14:41 PM »

In the experiment, we mix 0.5 ml Furan with a 4 molar sol'n of maleic anhydride in methylene chloride.

We know that the maleic anhydride is the limiting reagent but we are to show how many moles we are to add. Due to the fact that this was our first lab day, we were told to add ~2 to 2.5 ml of the anhydride solution.

I'm confused at finding the number of moles b/c of the chloride also added to it. My calculations don't give me the maleic anhydride as the limiting reagent. Where do I start?

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Borek

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Re: lmiting reagent calculation for maleic anhydride
« Reply #1 on: February 06, 2008, 12:21:46 PM »

Where do I start?

With balanced reaction equation.

Methylene chloride is just a solvent.
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phillyj

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Re: lmiting reagent calculation for maleic anhydride
« Reply #2 on: February 07, 2008, 10:56:14 AM »

ok, from what i can gather, it is a 1:1 mole ratio of Furan:Maleic Anhydride.

i guess since they gave us 0.5 ml furan we can try to find its moles. furan has a density of 0.936 g/ml so multiplying that with 0.5 ml gave 0.468 g furan. furan has MW=96 and [0.468g*(1mol/68g)= 0.006 mol furan

this is where i am confused. where do i use the 4 molars of Maleic Anhydride? and then what? my calculations give me furan as the limiting even though it is Maleic Anhydride

thanks
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Borek

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Re: lmiting reagent calculation for maleic anhydride
« Reply #3 on: February 07, 2008, 11:29:39 AM »

You know volume, you know concentration - you know everything required to calculate number of moles.
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phillyj

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Re: lmiting reagent calculation for maleic anhydride
« Reply #4 on: February 07, 2008, 11:44:09 AM »

actually for the anhydride, we are supposed to show how to get to the volume of 2 to 2.5 mls. given the 4 molar anhydride.
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Borek

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Re: lmiting reagent calculation for maleic anhydride
« Reply #5 on: February 07, 2008, 11:47:56 AM »

You've lost me... Anyway.

http://www.chembuddy.com/?left=concentration&right=molarity

C=n/V

Solve for unknown.
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phillyj

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Re: lmiting reagent calculation for maleic anhydride
« Reply #6 on: February 07, 2008, 12:05:18 PM »

this is why i'm confused.

basically we are given that 0.5 ml furan is needed for the diels alder rxn. and maleic anhydride is the limiting reagent. we were supposed to calculate the actual amount of the anhydride to use. the anhydride is 4 molars.

when i do it:
.5 ml furan*[density furan]* [MW furan] i get ~0.006 mol furan

i think thats right so far

next to calculate the amount of anhydride
0.006 mol furan*[1 mol anhydride/1 mol furan]= mols anhydride?

o get confused at the above part
 
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Borek

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Re: lmiting reagent calculation for maleic anhydride
« Reply #7 on: February 07, 2008, 12:18:28 PM »

If they react 1:1 you are OK. This is very basic stoichiometry.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
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phillyj

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Re: lmiting reagent calculation for maleic anhydride
« Reply #8 on: February 07, 2008, 12:31:34 PM »

so i will get 0.006 mols anhydride? that doesn't make sense to me.

how do i find the limiting reagent then?
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Borek

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Re: lmiting reagent calculation for maleic anhydride
« Reply #9 on: February 07, 2008, 12:36:32 PM »

You have to decide what you are trying to calculate. You either have KNOWN amounts of reactants and you want to know which one i s a limiting reagent, or you know amount of ONE reactant and you want to know how much of the other you need. You can't do both things at the same time.

Perhaps what you are looking for is the amount that will MAKE one of the ractants limiting reagent.
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phillyj

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Re: lmiting reagent calculation for maleic anhydride
« Reply #10 on: February 07, 2008, 01:00:19 PM »

yes, i think i am looking for the amount of anhydride that will make it the limiting reagent.

how do i do that?
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Borek

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Re: lmiting reagent calculation for maleic anhydride
« Reply #11 on: February 07, 2008, 09:23:04 PM »

Anything less then stoichiometric amount will do the trick.
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