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help please, excess, and % yeild - HSN forum

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help please, excess, and % yeild


5 replies to this topic

#1 lauraangel

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Posted 28 May 2007 - 01:10 PM

if u have pp, it would be good...

2002 writeen paper Q.8.a)(ii)
ave dun something n got the right answer but not sure if a just dun somthin n then guessed lol

and also 10.(d) % yeild these arwnt bad wen u no how, but unfortunetelly i dont no how:(


thanks xxxxx

#2 BuckminsterFullerene

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Posted 28 May 2007 - 01:26 PM

Hi Laura,
Well firstly in any excess calculation you must tabulate respective numbers of moles in each reactant.

Firstly, Silver Nitrate.

Divide its mass by its Relative Formula Mass, so 0.2/169.9 and you get 0.0011

Next reactant, Hydrochloric acid, Number of moles = Concentration x Volume
= 0.001 x 0.02 (volume in litres) = 0.00002

Now, you must look at the Reactant Ratio, just now there are no numbers preceeding each chemical formula hence, the incumbent ratio is 1:1
so
You have 0.0011 and 0.00002 and both need to react in a one to one ratio, so you only need 0.00002 moles of Silver Nitrate, so it is in excess.
Chemical Engineering - Edinburgh

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#3 lauraangel

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Posted 28 May 2007 - 01:38 PM

ok thanks, i dont think i will be able to get the hang of these b4 2moro sad.gif any1 help with the % yeild?

#4 BuckminsterFullerene

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Posted 28 May 2007 - 01:52 PM

I can't be arsed though I'll still help!

You must compare the relative formula masses of both the reactant you are considering and the product who's yield you are approximating.

Hence 3H2 ---> 2NH3
GFM's 6 ---> 34
Have, 200 ---> x (You need to know what this will give you according to the previous ratio)

so x = 34 x 200 all divided by 6
so x = 1133 and this is what you should be getting when you use 200kg of Hydrogen

Though this is only theoretical and you actually get 650

so Percentage yield = 650/1133 x 100 = 57.36%
Chemical Engineering - Edinburgh

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#5 lauraangel

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Posted 28 May 2007 - 02:00 PM

ww thnk so much, u little cant be arsed kind person u, lol xxx

#6 Cooldude

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Posted 28 May 2007 - 02:12 PM

here is another way to do the percentage yield one, i personally prefer it, using more equations as opposed to ratios...

first write out the balanced equation:

3H2 + N2--------------> 2NH3

find out the number of moles of hydrogwn, since you are given the mass of hydrogen used:

n=m/GFM
=200/2
= 100mol oh hydrogen

there are 3 moles of hydrogen for every 2 moles of ammonia
therefore for 100 moles of hydrogen, there are 66.667 moles of ammonia

then m=nxGFM
= 66.667x17
= 1133kg- this is the theoretical yield of ammonia

percentage yield= actual yield/theoretical yield x 100
= 650/1133 x 100
= 57.4%

i don't know if you'll like this way but i just prefer it, to use the set equations smile.gif





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