posted by Johnny on .
When 100mL of 1.00 M Ba (NO3)2 and 100 mL of 1.00 M Na2SO4, both at 25 degrees celsius, are mixed in a constant pressure calorimeter, a white precipitate is formed and the temperature rises to 28.1 degrees celsius.
A. What is the precipitate?
B. If all of the heat is absorbed by the solvent (water), what is the molar enthalpy change for the reaction.
(s for water = 4.18 J/g.C)
I really want to understand this problem so if someone could show me step by step I would really appreciate it so much!
A. Here is a simplified list of the solubility rules. You can find the material that ppts by studying this list.
B. How much heat is absorbed by the water? That is q = heat absorbed = mass H2O x specific heat water x delta T. How many moles of the ppt do you have. (hint: Write the equation an balance it to find moles ppt). Then deltaH q/mol ppt.
Post your work if you get stuck.
So for part A i balanced the equation to:
Ba(NO3)2 + Na2 -> Ba(SO4) + 2Na(NO3)
then I got: (NO3)2 + Na2 -> 2NaNO3 does that mean 2NaNO3 is the precipitate?
For part B I got q= 233.5 J
and now I'm stuck lol
A. No, it doesn't mean NaNO3 is the ppt. You didn't read the table in the link I gave you. That link tells you that sulfates are soluble EXCEPT for (see #4 in the list).
Ba(NO3)2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaNO3(aq).
B. 233.5 is not right.I don't know what show your work means to you but it doesn't mean to show the answer only. I'm headed to bed and I don't have time to wait around while we go through this so the problem follows:
mass x specific heat x delta T
200 g H2O x 4.18 x (28.1-25) = ?? = q
delta H/mol = q/mol = q/0.1 = ??