7 Responses

1. sirong

   26|Sep|2007 1

   is the as D ? cos any gases at r.t.p is always 24 dm^3 ? am i right ?

   Reply

2. platinumwinter

   27|Sep|2007 2

   amt of aluminum liberated=9/27
   amt of electrons=9/27*3=1 mol (because Al3+)
   4OH- => 2H2O + 2O2 + 4e-
   1 mol e- means 0.5mol O2 means 0.5*24=12dm3 O2 liberated so ans is
   C, 12dm^3
   isit correct? someone pls verify

   Reply

3. robobooga

   29|Sep|2007 3

   anyone can give the answer for this qn?

   Reply

4. Sean Chua

   29|Sep|2007 4

   Hi Sirong and Platinum,

   Both answers are wrong. This MCQ question requires us to think of a few more steps to get the answer. In the question, it involves Electrolysis, where Al metal is formed from Al3+ ions, and 02 gas is formed from the dilute sulphuric acid present.

   First we need to look at the two half equations:

   Al3+(aq) + 3e- => Al(s) —- (1)

   from dilute H2SO4, 02 gas will be liberated by the OH- ions present in water (dilute acid contains water), and equation is:
   4OH-(aq)=>O2(g) + 2H20(l) + 4e- –(2)

   Multiply Eq (1) by 4, and Eq (2) by 3, this will allow the electrons to be the same, and allows us to combine to form overall equation of:

   4Al3+ + 12OH- + 12e- => 4Al + 3O2 + 6H20 + 12e-

   Mole of Al liberated=9/27=1/3 mole
   Based on overall equation, 4 mole of Al will gives 3 moles of O2.
   Thus, 1/3 mole of Al will give 1/4mole of of O2.
   Vol of O2 liberated = 0.25mole x 24dm3/mole = 6dm3

   Ans: A

   Hope this is clear.
   Sean
   SimpleChemConcepts

   Reply

5. Sean Chua

   30|Sep|2007 5

   Hope you all understand the working to the answer for this question. Students tends to skip the steps on working out the half equations for the cathode and anode reactions of the electrolysis. Rmbr to use both half reactions and then get the overall reactions. It is only then you can use mole concept to solve the question. Difficulty? Not too bad. But you need to be exposed to such question in case it comes out in this year O Level.

   Sean
   SimpleChemConcepts

   Reply

6. waleed bin qasim

   10|Oct|2009 6

   welldone sir u r gr8 teacher. Allah would give u suffice blessings

   Reply

7. Saud

   27|Feb|2010 7

   A
   6dm cubic
   finding moles of aluminium that make up to 0.333 as 9/27=0.333moles,
   so,
   as the charge on an aluminium ion is +3 so three moles of electricity i-e electrons is used to discharge it,
   this makes the moles of electricity to be 3 X 0.333 = 0.999moles,,

   now taking the oxygen ion, it has a charge of -2, i-e, one mole of an oxygen ion gives 2 mole of electrons/electricity……………

   so, 0.999 moles of electrons get liberated from how much oxygen ions??…………

   this makes it to be 0.999/2 = 0.4995 moles of oxygen ions………

   one ion of oxygen is written as O(-2 ) and one mole of oxygen is written a O(2)……………………..so 0.4995 moles of oxygen will liberate 0.4995/2=0.24975moles of oxygen gas…………………..

   corresponding to it, the volume of oxygen gas liberated at the anode will be as M.G.V / moles of oxygen,

   i-e,

   24 X 0.24975=5.994dm cubic which rounding off makes as 6dm cubic…………………………. hope im not wrong in my calculations

   Reply