Calcium;sulfate CaO4S CID 522675 - structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities, safety.
The dissolution of calcium chloride in water
CaCl2(s)--> Ca^2+ (aq) +2 Cl^-1 (aq)
A) Use the data from appendix C. and below to determine Delta Hof formation in kj/mol for this process (deltaH(Ca^2+(aq)=-542.8kj/mol)
![Balanced Dissolution Of Caso4 Balanced Dissolution Of Caso4](https://researcherridge2014.files.wordpress.com/2014/10/justine-sauvage.jpg)
so what I basically understand is were ment to use an equationthat delta H= sum of products- sum of reactants
..and that would be something like this =[(Ca^2+)+2(Cl-)]-[CaCl2]
[-542.8+(2x?)]-[-795.8]
I cant find the delta H for the Cl-, and even if I did, I wasunder the impression that multiplying..like the Cl- by its 2 moleswould cancel out the moles from your kj/mol and you would just beleft with kj.. and does it still work like that if there is onlyfor instance the one mole of CaCl2?
Am I on the right track even? Please show work if you can
and im not sure if part B is important to part A or not but I'llput it anyway just in case
B) suppose that 20 g of CaCl2 is dissolved in 0.1 L of water at20 degrees C. Calculate the temperature reacher by the solution,assuming it to be an ideal solution with a heat capacity close tothat of 100g of pure water (418 J/degrees C)
1 Answer
Nov 7, 2015
Answer:
Explanation:
First tally the atoms. Since #SO_4^'2-'# is an ion, I'm going to consider it as one 'atom' in order to not confuse myself.
Based on the subscripts,
left side:
#Ca# = 1
(#SO_4# ) = 1
#Al# = 1
#Cl# = 3
(
right side:
#Ca# = 1
(#SO_4# ) = 3
#Al# = 2
#Cl# = 2
(
Let's start balancing the most complicated 'atom', #SO_4^'2-'#
left side:
#Ca# = (1 x #color (red) 3# ) = 3
(#SO_4# ) = (1 x #color (red) 3# ) = 3
#Al# = 1
#Cl# = 3
(
right side:
#Ca# = 1
(#SO_4# ) = 3
#Al# = 2
#Cl# = 2
(
Since #CaSO_4# is a substance, we need to also multiply the coefficient with its #Ca# atom. Now that there are 3 #Ca# atoms on the left, there must be 3 #Ca# atoms on the right.
left side:
#Ca# = (1 x 3) = 3
(#SO_4# ) = (1 x 3) = 3
#Al# = 1
#Cl# = 3
(
right side:
#Ca# = (1 x #color (blue) 3# ) = 3
(#SO_4# ) = 3
#Al# = 2
#Cl# = (2 x #color (blue) 3# ) = 6
(
Again notice that since #CaCl_2# is a substance, the coefficient 3 should also be applied to its #Cl# atoms. Since there are 6 atoms of #Cl# on the right, we need to also have the same number of #Cl# atoms on the left.
left side:
#Ca# = (1 x 3) = 3
(#SO_4# ) = (1 x 3) = 3
#Al# = (1 x #color (green) 2# ) = 2
#Cl# = (3 x #color (green) 2# ) = 6
(
right side:
#Ca# = (1 x 3) = 3
(#SO_4# ) = 3
#Al# = 2
#Cl# = (2 x 3) = 6
(
![Calculate the solubility of calcium sulfate in water Calculate the solubility of calcium sulfate in water](/uploads/1/2/3/9/123916928/375215464.jpg)
Again, since #AlCl_3# is a substance, the coefficient should also apply to the bonded #Al# atom.
Now the equation is balanced.
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