This problem's been confusing the heck out of me. "Consider the reaction: 2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq)+2H2O(l) deltaH= -118 kJ Calculate the heat when 100.0 mL of 0.5 M HCl is mixed with 300.0 mL of 0.5 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25 C and that the final mixture has a mass of 400.0 g and a specific heat capacity of 4.18 J/C*g, calculate the final temperature of the mixture." There is no answer given to me for this problem so I want to make sure if I am doing this right, which I doubt I am. The net reaction is H2 + 2(OH) -> 2H2O deltaH= -118 The reason I chose H2 as opposed to 2H is because that's how H occurs in its natural state, and it would still respect the initial molar quantities. Now I found the limiting reagent. I know I have 0.05 mols of H2 and .15 mols OH, and if I divide .15 by 2mols OH then .075 > .05 so H2 is the limiting reagent. I then find out how many mols of H2O this would give me, which would be 0.05 * 2/1 = 0.1 mols H2O. Now I find the kJ/mol H2O enthalpy, which is -118/2 = -59 kJ/mol H2O. Multiplying that by .1 gives me -5.9 kJ. Since the reaction is losing energy, the energy perceived by the calorimeter is positive and so there will be an increase in temperature. So 5900 J = 4.18 * 400 * (x-25), x = 28.52871 C. Is this right? Thanks.