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ALL CIE CHEMISTRY DOUBTS HERE !!

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Amelia:

--- Quote from: Sue T on January 19, 2011, 07:17:42 am ---i have an A2 ppr 5 ques
so we have this eqn:
CuCO3.Cu(OH)2.x H2O(s) ------> 2CuO(s) + CO2(g) + (x+1)H2O(g)
we're tryin 2 find x
wat i got is:
1) the mass of the basic copper carbonate (LHS) and;
2) loss in mass which is basically CO2 and H2O
there wer 2 expressions in the m.s. tht im supposed 2 use:
(221 + 18x); 44 + 18(x + 1);
ny has ny idea wat 2 do 2 find x ??
p.s. i hate paper 5. >:(

--- End quote ---

eg :- mass of empty boiling-tube - 10.32g
mass of boiling-tube and basic carbonate before heating - 11.19g
mass of boiling-tube and residue after heating - 10.92 g.

After the heating the CO2 and H2O boil away and only CuO remains.
(11.19 - 10.92 g)(mol / 159.09 g) = 1.70e-3 mol rxn

(1.70e-3 mol)(221.1138 + 18.0148x) = 11.19 - 10.32 g
x = 16 (rounding to 2 s.d.)

P.S - I hope to be right.  ::)

Sue T:
 :-[ ummm... im sorry - i dont seem to get the working
IF there is no bother - pls elaborate just a lil bit more - im sure im jus missin smthin somewere
please and thank you  :-\

Amelia:

--- Quote from: Sue T on January 19, 2011, 07:17:42 am ---After the heating the CO2 and H2O boil away and only CuO remains.
(11.19 - 10.92 g)(mol / 159.09 g) = 1.70e-3 mol rxn

--- End quote ---

11.19 - 10.92 = 0.27g (mass of residue)

CuO - (63.546+16)x2

Find the moles.
The moles of the residue - 0.87/((63.546+16)x2) = 1.70e-3 mol rxn



Moles x RMM = Mass.
Therefore, use this eqn. to find X.
(1.70e-3 mol)(221.1138 + 18.0148x) = 11.19 - 10.32 g

Sue T:
got it  ;) thank you - sorry 4 th trouble  :)

thecandydoll:
Heyy!
I need resources and help in Chemistry P5  Got no idea.No IDEA how to attempt this paper.

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