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#1
Rep for a guided answer - even though its not worth much- id reall appreciate it thank you! can i see the working out in terms of X and Y and the 121, thank you!

KHCO3 (S) + HCL (AQ) ---> KCL (AQA) + H20 (L) + CO2 (G) LET THE ENTHALPY IS X FOR THIS
MgCo3 (s) + 2HCL (aqa) ---> MgCl2 (aq) + H20 (L) + CO2 (g) LET THE ENTHALPY BE Y FOR THIS

KHCO3 (S) ----> KHCO3 (AQA) = + 121 KJ MOL

CALCULATE THE ENTHALPY CHANGE FOR THE FOLLOWING BY CONSUTRCTUING A HESS LAW CYCLE.
MgCl2 (aqa) + 2KHCO3 (aqa) ---> MgCo3 + 2KCl (aqa) + H20 (l) + CO2 (g)
0
14 years ago
#2
(Original post by LondonBoy)
Rep for a guided answer - even though its not worth much- id reall appreciate it thank you! can i see the working out in terms of X and Y and the 121, thank you!

KHCO3 (S) + HCL (AQ) ---> KCL (AQA) + H20 (L) + CO2 (G) LET THE ENTHALPY IS X FOR THIS
MgCo3 (s) + 2HCL (aqa) ---> MgCl2 (aq) + H20 (L) + CO2 (g) LET THE ENTHALPY BE Y FOR THIS

KHCO3 (S) ----> KHCO3 (AQA) = + 121 KJ MOL

CALCULATE THE ENTHALPY CHANGE FOR THE FOLLOWING BY CONSUTRCTUING A HESS LAW CYCLE.
MgCl2 (aqa) + 2KHCO3 (aqa) ---> MgCo3 + 2KCl (aqa) + H20 (l) + CO2 (g)
This looks very similar to the question I answered further down the page - again by you. Why don't you check that one out.
0
14 years ago
#3
Rewrite that like that:
2KHCO3(aq) -----> 2KHCO3(s) ................................ ........ 2dH1 = (-121 x 2)kJ/mol
2KHCO3 (s) + 2HCL (aq) ---> 2KCL (aq) +2H2O (l) + 2CO2 (g)......2dH2 = 2X
MgCl2 (aq) + H2O (l) + CO2 (g) ----> MgCO3(s) + 2HCl(aq) ........dH3 = -Y
--------------------------------------------------------------
MgCl2(aq) + 2KHCO3(aq) ---> MgCO3(aq) + 2KCl(aq) + H2O(l) + CO2(g)
so dH = 2dH1 + 2dH2 + dH3
= (-121 x 2) + 2X - Y

Hope you can do all of this type of question
Cheers
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