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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Sign up to join this communityUsing 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?
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.