Measure total solution mass and ΔT; compute solution heat (and calibrated cup heat when supplied); use an energy balance to obtain reaction heat; then divide by the amount corresponding to the requested molar basis. For ν mol of limiting reactant per mole of reaction as written, ξ = n(limiting)/ν and ΔrH = q(reaction)/ξ. For 2 H₂ + O₂ → 2 H₂O, reacting 0.20 mol H₂ corresponds to 0.10 mol of reaction, not 0.20 mol. Dividing by moles of H₂ instead reports heat per mole H₂.
Suppose 50.0 mL of 1.00 M HCl reacts with 40.0 mL of 1.00 M NaOH. The supplied amounts are 0.0500 mol H⁺ and 0.0400 mol OH⁻, so 0.0400 mol water forms and acid remains in excess. With additive volumes, density 1.00 g/mL, c = 4.18 J/(g·°C), ΔT = 6.00 °C and Ccal = 10.0 J/°C, qsurroundings = (90.0)(4.18)(6.00) + (10.0)(6.00) = 2317.2 J. Thus qreaction = −2.3172 kJ and ΔH = −2.3172/0.0400 = −57.9 kJ/mol water. Dividing by the excess 0.0500 mol acid would assign heat to acid that did not react.