A student mixed 75.0 mL of water containing 0.75 mol HCI at 25* C with 75.0 mL of water containing 0.75 mol of NaOH at 25*C in a foam cup calorimeter. The temperature of the resulting solution increased to 35*C. How much heat in kilojoules was released by this reaction?

C (water) + 4.18 J/g X C

q=mass water x specific heat water x delta T.

mass H2O = 150 g

To calculate the heat released by this reaction, we need to use the equation:

q = m * C * ΔT

where q is the heat released or absorbed (in joules), m is the mass of the water (in grams), C is the specific heat capacity of water (in J/g * °C), and ΔT is the change in temperature (in °C).

First, calculate the mass of water in the mixture. Since both solutions have equal volumes of 75.0 mL, the total volume is 150 mL or 150 grams since the density of water is approximately 1 g/mL.

Next, calculate the moles of HCl and NaOH in the mixture. Given that both solutions have 0.75 mol each, the total moles in the mixture are 1.5 mol.

Assuming that the reaction between HCl and NaOH is complete, we can use the fact that the enthalpy change (ΔH) for the neutralization of 1 mole of HCl with 1 mole of NaOH is -57 kJ/mol.

Now, calculate the heat released by the reaction:

q = ΔH * n

where ΔH is the enthalpy change and n is the number of moles.

q = -57 kJ/mol * 1.5 mol = -85.5 kJ

Therefore, the heat released by this reaction is -85.5 kilojoules (kJ). The negative sign indicates that the reaction is exothermic, meaning heat is released.