Is there anybody that can tell me if I'm doing this question right? Metallic Ca reacts with hydrochloric acid according to the following equation: Ca(s) + 2 HCl(aq) ---> 2 CaCl2 (aq) + H2(g) The enthalpy change associated with this reaction can be determined with an ice calorimeter. If 13.59 g of ice are melted when 0.3487 g of Ca reacts with excess hydrochloric acid, what is deltaH (in kJ/mol Ca) for the reaction written above? The heat of fusion of ice is 333 J/g. qrxn + qfusion = 0 qrxn = ndeltaHrxn =(13.59 g H2O)(0.333 kJ/g) =4.53 kJ 0.3487 g Ca X 1 mol Ca/40.08 g Ca = 8.70 x 10^-3 mol 4.53 KJ/8.70 x 10^-3 mol = 521 kJ/mol? Does this look right?