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

**0**

# help please, excess, and % yeild

Started by lauraangel, May 28 2007 01:10 PM

5 replies to this topic

### #1

Posted 28 May 2007 - 01:10 PM

### #2

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.

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

*Conditional*### #3

Posted 28 May 2007 - 01:38 PM

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

### #4

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%

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

*Conditional*### #5

Posted 28 May 2007 - 02:00 PM

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

### #6

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

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

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