1. The problem statement, all variables and given/known data When 50.0 mL of 1.0 M H2SO4 at 26.1 celcius was added to 50.0 mL of .976 M NaOH also at 26.1 celcius, the temperature rose to 32.6 celcius. Calculate the heat of reaction. 2. Relevant equations H2SO4 (aq) + 2NaOH(aq) --> Na2SO4 (aq) + 2H2O(l) 3. The attempt at a solution First, I converted both given amount of solutions into moles and found that NaOH was the limiting reactant (made .0488 mol h20 while the sulfuric acid made .1 mol). Then I converted the 50 mL of NaOH into grams (1 g/mL) and plugged that into the q=mcDeltaT equation. I did: 50g*4.184J/gcelcius*6.5 celcius = 1359.8 J Then I divided that by the mols of NaOH 1359.8 J/.0488 mol = 27864.7541 J/mol Can anyone confirm if I did this correctly?