Ok, Im stuck half way through this problem. heres what i got: a hyopthetical substance A_2 reacts with B_2: 3 A_2 + 4 B_2 --> 2 A_3 B_4 delta H = ??? kcal/mole A_2 1 mole of A = 56.4g 1 mole of B = 29.6g when 15.6g of A_2 and an excess of B_2 react in a calorimeter that contains 0.00186 mL of water, the water temp changes from 23/5 celsius to 86.7 celsius. What is the delta H of the reaction in kcal/mole? I approached this by first solving for q , then divide it by mole. But im not sure if i solved for the mole correctly because i multiplied 56.4g by 6 since its "3 A_2". Also, after i get the answer, am i suppose to divide it by 3 because in the reaction, the coefficient for A_2 was 3, that doesn't match with what the answer wanted, which is kcal/ mole A_2. (coefficient is 1) Any help?