How much heat is released when 22.12 grams of solid calcium hydroxide is added to 334 mL of 6.4 M HNO3?

Question

I TRIED SOLVING IT BUT I DONT KNOW IF AM RIGHT

Ca (OH) 2 (S) + HNO3 (aq)>>>>> 2H2O (l) + Ca (NO3)2 (aq)

Ca (OH) 2 (S)=-986.1 kj/mole
+
2HNO3 (aq) = -414.8 kj/mole
= -1400.9 kj/mole

2H2O (l)= 571.6 kj/mole
+
Ca (NO3)2 (aq) = -938.4 kj/mole
=-1510

-1510-(-1400.9)= -109.1




the mole for Ca(OH)2 (s)= .2985450115

and the moles for HNO3=2.1376

Answer 1

Did you multiply the heat of formation of HNO3 and of H2O by 2? You need to do that since you have 2 moles of each of those substances in the reaction.

If you did do that, you're almost finished. The answer you've calculated so far (-109.1) is the heat (in kilojoules) released by reacting 1 mole of Ca(OH)2 with the HNO3.

Just multiply that number by the number of moles of Ca(OH)2 that you've used, and you should get to the final answer, which will have units of just kilojoules.