In a coffee cup calorimeter, 50 mL of 0.1 M HCl and 50 mL of 0.1 M NaOH are mixed to

yield the following reaction: HCl + NaOH  NaCl + H2O. The two solutions were initially at 22.1 oC and the final temperature is 24.2 oC. Calculate the heat that accompanies this reaction in kJ/mol of H2O formed. Assumed the combined solution has a mass of 100 g and a specific heat of 4.18 J/gK

How many mols H2O were formed? That's 50 x 0.1 = 5.0 millimols or 0.005 mols.

q = heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial)
That is q for 0.05 mols. Convert to J/1 mol and to kJ/mol.

8.778 KJ/Mole

To calculate the heat that accompanies this reaction, you can use the equation:

q = m * c * ΔT

Where:
q is the heat transferred in joules (J)
m is the mass of the solution in grams (g)
c is the specific heat capacity of the solution in J/gK
ΔT is the change in temperature in Kelvin (K)

First, let's calculate the mass of the combined solution:
The volume of the combined solution is 50mL + 50mL = 100mL = 100g (since 1mL of water is approximately equal to 1g)

Next, calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 24.2°C - 22.1°C = 2.1°C

Since the specific heat capacity (c) is given as 4.18 J/gK, we can now calculate the heat transferred (q):

q = 100g * 4.18 J/gK * 2.1K

Now we need to convert the heat from joules to kilojoules:

q = (100g * 4.18 J/gK * 2.1K) / 1000

Finally, to calculate the heat in kJ/mol of H2O formed, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that 1 mol of H2O is formed for every 1 mol of NaOH reacted.

Thus, the heat in kJ/mol of H2O formed is the same as the heat transferred (q) calculated above in kilojoules.

Please plug in the values into the equation and calculate the final answer.