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?
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$\begingroup$ Are you sure you have the right units for $\Delta H$? $\endgroup$ – LDC3 Apr 20 '14 at 20:54
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$\begingroup$ @ldc3 Ya, it was a given value for the problem. $\endgroup$ – EmptyStuff Apr 20 '14 at 20:58
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$\begingroup$ 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
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$\begingroup$ $\Delta H$ 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
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$\begingroup$ @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
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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.
