Qualification > Sciences
CHemistyr :Excess Reagent!!
JD46:
I had mentioned earlier , though my posts were removed 4 some reason,it evades it why? but in case of any violation, my intention was innocent and i was desperate, i mean who wouldn't be wen you have your boards in couple of days.
So how do i identify an excess reagent!? i don't know how to decide as per the number of moles? its so wierd!,thanx guys! :)
elemis:
What do you mean your posts were removed ? As in deleted ?
nid404:
There must've been a reason. I don't recollect removing any posts though...so I am not aware.
.could you post some question...so it'll be easier for me to explain :)
Ivo:
--- Quote from: JD46 on June 04, 2010, 03:50:37 pm ---I had mentioned earlier , though my posts were removed 4 some reason,it evades it why? but in case of any violation, my intention was innocent and i was desperate, i mean who wouldn't be wen you have your boards in couple of days.
So how do i identify an excess reagent!? i don't know how to decide as per the number of moles? its so wierd!,thanx guys! :)
--- End quote ---
Look at the last 'example' here, if you can understand that, then done! Very simple! ;)
It's hard to 'teach' you moles. My advice is go through past paper questions, and then you'll find it is well easy!
I've written very thorough explanations for some questions and doubts people have asked. Here they are:
s02, Q5) c) and d):
--- Quote from: Ivo on May 30, 2010, 01:00:49 pm ---The equation you need to use is:
Volume of gas (in dm3)
Moles of gas = --------------------------
24
Note to convert cm3 to dm3, divide quantity by 1000 (e.g. 20cm3/1000=0.02dm3)
This is the proper theory and equation behind my calculations below.
Here are the answers for 5) c) as requested:
i) Moles of C4H6: 0.02/24=0.000833
From equation, we know 2 moles of C4H6 reacts with 11 moles of O2
Therefore, moles of O2: 0.000833*5.5=0.00458
Therefore, volume of O2: 4.58*24*1000=110cm3
ii) From equation, we know 2 moles of C4H6 produces 8 moles of CO2
Therefore, moles of CO2: 0.000833*4=0.00333
Therefore, volume of CO2: 0.00333*24*1000=80cm3
iii) From equation, we know 2 moles of C4H6 produces 6 moles of H2O
Therefore, moles of H2O: 0.000833*3=0.0025
Therefore, volume of H2O: 0.0025*24*1000=60
Therefore, total volume of gases is simply volume of CO2 + volume of H2O:
80+60=140cm3
Alternatively, 8+6=14, so 2 moles of C4H6 produces 14 moles of products
Therefore, moles of products: 0.000833*7=0.00583
Therefore, volume of products: 0.00583*24*1000=140cm3
For part d), you use this formula:
Mass
Moles = ------
Mr
d) From the above equation, 1 mole of butyne forms 3 moles of water
Number of moles of butyne reacted: 9/54=0.167
Number of moles of water formed: 0.167*3=0.5
Mass of water formed: 0.5*(2+16)=9g
I hope this has helped. :D
--- End quote ---
s08, Q7) b)
--- Quote from: Ivo on May 31, 2010, 11:38:26 am ---OK, hopefully this clears your doubt.
This time, you'll need to also apply this formula:
Concentration (mol/dm3) = Moles / Volume (dm3)
So:
i) Number of moles of NaOH used = 2.24*0.025 = 0.056 mols
ii) Maximum number of moles of Na2SO4.10H2O that could be formed = From the equation given, you can see 2 moles of NaOH gives 1 mole of Na2SO4.10H2O. So it is simply: 0.056/2 = 0.028 mols
iii) Mass of one mole of Na2SO4.10H2O = 322g
iv) Maximum yield of sodium sulphate-10-water = 0.028*322 = 9.02g
v) Percentage yield = (3.86/9.02)*100 = 42.8%
If you don't understand any of this, I'll be happy to explain ;)
--- End quote ---
w08, Q7) a)
--- Quote from: Ivo on May 30, 2010, 01:00:49 pm ---The equation you need to use is:
Volume of gas (in dm3)
Moles of gas = --------------------------
24
Note to convert cm3 to dm3, divide quantity by 1000 (e.g. 20cm3/1000=0.02dm3)
This is the proper theory and equation behind my calculations below.
Here are the answers for 5) c) as requested:
i) Moles of C4H6: 0.02/24=0.000833
From equation, we know 2 moles of C4H6 reacts with 11 moles of O2
Therefore, moles of O2: 0.000833*5.5=0.00458
Therefore, volume of O2: 4.58*24*1000=110cm3
ii) From equation, we know 2 moles of C4H6 produces 8 moles of CO2
Therefore, moles of CO2: 0.000833*4=0.00333
Therefore, volume of CO2: 0.00333*24*1000=80cm3
iii) From equation, we know 2 moles of C4H6 produces 6 moles of H2O
Therefore, moles of H2O: 0.000833*3=0.0025
Therefore, volume of H2O: 0.0025*24*1000=60
Therefore, total volume of gases is simply volume of CO2 + volume of H2O:
80+60=140cm3
Alternatively, 8+6=14, so 2 moles of C4H6 produces 14 moles of products
Therefore, moles of products: 0.000833*7=0.00583
Therefore, volume of products: 0.00583*24*1000=140cm3
For part d), you use this formula:
Mass
Moles = ------
Mr
d) From the above equation, 1 mole of butyne forms 3 moles of water
Number of moles of butyne reacted: 9/54=0.167
Number of moles of water formed: 0.167*3=0.5
Mass of water formed: 0.5*(2+16)=9g
I hope this has helped. :D
--- End quote ---
s04, Q7) b):
--- Quote from: Ivo on June 04, 2010, 12:40:43 pm ---7) b) i) Moles of Mg = 3/24 = 0.125
Moles of CH3COOH = 12/60 = 0.200
Using ratios given in the equation, to compare, divide 0.2/2 = 0.100 moles of CH3COOH.
Therefore, magnesium is in excess.
ii) The one not in excess: ie. 0.1
iii) 0.1*24 =2.4dm3
Got it?
--- End quote ---
I hope that's enough questions for you to follow! ;P
JD46:
okie dokie, here a question nid
1)3.0g of magnesium was added to 12.0g of ethanoic acid.
Mg + 2CH3COOH ? (CH3COO)2Mg + H2
The mass of one mole of Mg is 24 g.
The mass of one mole of CH3COOH is 60 g.
(i) Which one, magnesium or ethanoic acid, is in excess? You must show your
reasoning.[3] MJ04
P.S.I got some more doubts hope ya dont mind! ;D
-How to do i draw the humongous struucture of Sio2
-And how would i draw the polymer of the following structure? the picture is attached.
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