The enthalpy change when a strong acid is neutralized by strong base is -56.1 kj/mol. if 135 ml of 0.450 m hi at 23.15 °c is mixed with 145 ml of 0.500 m naoh, also at 23.15 °c, what will the maximum temperature reached by the resulting solution? [assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 j/g·c, and that the density of the final solution is that of water.]

26.06°C First, determine the number of moles of each reactant to determine the limiting reactant. You do this by multiplying the volume (in liters) by the molarity. So Moles HI = 0.135 * 0.450 = 0.06075 mol Moles NaOH = 0.145 * 0.5 = 0.0725 mol So the limiting reactant is HI and the number of moles is 0.06075. Now determine how many kilojoules are released by multiplying. 0.06075 mol * -56.1 kJ/mol = -3.408075 kJ Since the enthalpy is negative, the reaction is exothermic and will increase the temperature of the solution. So we now have 0.135 + 0.145 = 0.28 l =280 ml of a solution with a density of 1 g/ml and a specific heat of 4.18 J/g·C and we need to determine how much the temperature will rise with the addition of 3408 J of energy. So 280 ml * 1 g/ml = 280 g 280 g * 4.18 J/g·C = 1170.4 J/C 3408.075 J / 1170.4 J/C = 2.911889 C So the temperature will rise by 2.91 °C. Add that to the initial temperature, getting 23.15 °C + 2.91 °C = 26.06°C Maximum temperature reached will be 26.06°C

You need to divided twenty four by a dozen them by the temperature it is getting cook by

Icant see the pictures