0
$\begingroup$

Using this equation, $\ce{CaCl2(s) -> Ca^{2+}(aq) + 2Cl^{-}(aq)}$

and $\Delta H = - 325.0 \; kJ $

If you have 15.0 grams of the solid, how much heat is released or gained in kilojoules?

enter image description here

|improve this question
$\endgroup$
  • $\begingroup$ Are you sure you have the right units for $\Delta H$? $\endgroup$ – LDC3 Apr 20 '14 at 20:54
  • $\begingroup$ @ldc3 Ya, it was a given value for the problem. $\endgroup$ – EmptyStuff Apr 20 '14 at 20:58
  • $\begingroup$ Suppose you dissolve a gram of calcium chloride and then a kilogram, both in excess water. They cannot each and both be the value you posted. If you have the correct units, the path to a correct answer is obvious. Open your textbook. The first time you do it there is a crackling sound. Google it, "enthalpy of solution". BTW, the value is for infinite dilution. $\endgroup$ – Uncle Al Apr 20 '14 at 21:03
  • $\begingroup$ $\Delta H$ is usually in $kJ/mol$. That is why I am asking. Either the question has a mistake or maybe it is $\Delta H^°$ $\endgroup$ – LDC3 Apr 20 '14 at 21:03
  • $\begingroup$ @LDC3 I updated the post with a picture of the question. It very may well be kJ/mol but it isn't stated. $\endgroup$ – EmptyStuff Apr 20 '14 at 23:06
1
$\begingroup$

I'm going to assume that the question is presenting the given change in enthalpy as a molar enthalpy change (so as to say that -325.0 kJ are evolved per mole of CaCl2).

Under that assumption, answering the question is then a matter of using the a molar fraction of CaCl2 for the amount you have present.

CaCl2 has a molar mass of 110.98 g/mol so with 15.0 g of CaCl2, you'd have roughly 0.1352 mol of CaCl2. 0.1352 mol * -325.0 kJ/mol = -43.94 kJ. The negative here means that the CaCl2 will release 43.94 kJ of heat.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.