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]

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!

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):

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 = ------
               M_r

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.

s08, Q7) b)

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

w08, Q7) a)

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 = ------
               M_r

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.

s04, Q7) b):

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?

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!
-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.

[J.Darren]

https://studentforums.biz/index.php/topic,8469.msg251999.html#msg251999

1 Si is bonded to 4O
1 O is bonded to 2Si

H  COOCH3
  C
H  H

[Ivo]

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!
-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.

Most of your doubts has been answered before:

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.

Got it?

For Silicon Dioxide, see below.

[nid404]

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

Mole Ratio of Mg to CH3COOH is 1:2
3g of Mg= 3/24 =0.125 moles
so theoretically moles of CH3COOH should be twice = 0.25
12g of CH3COOH= 12/60= 1/5 moles=0.2 moles
So CH3COOH is lesser than the theoretical value. Hence Mg is in excess

[JD46]

Thanks a ton guys for the over whelming response!

-By the way, how do i polymerise the structure
-and the ususal, naming the blessed alkane's,alkene's isotopes, and  forming their structure and  drawing,..basically im  in structural problems!! 

Thanks  frnds!!

[elemis]

As all IGCSE chemistry exams are over, this topic will be locked.