Suppose 100.0 mL of 1.00 M HCl and 100.0 mL of 1.00 M NaOH, both initially at 23.0°C, are mixed in a thermos flask. When the reaction is complete, the temperature is 29.8°C. Assuming that the solutions have the same heat capacity as pure water, compute the heat released (in kJ).

........HCl + NaOH ==> NaCl + H2O

mols HCl = 0.1 x 1.00 = 0.1
mols NaOH = 0.1 x 1.00 = 0.1
mols H2O produced = 0.1
q = mass x specific heat H2O x (Tfinal-Tinitial)
q = 200 x 4.184 x (29.8-23.0)
That gives q for 0.1 mol in J. Convert to kJ/mol if desired.

To compute the heat released in this reaction, we can use the formula:

q = m * C * ΔT

where
q is the heat released (in joules),
m is the mass of the solution (in grams),
C is the specific heat capacity of water (4.18 J/g·°C), and
ΔT is the change in temperature (in °C).

To find the mass of the solution, we can use the formula:

m = V * ρ

where
m is the mass of the solution (in grams),
V is the volume of the solution (in milliliters), and
ρ is the density of water (1.00 g/mL).

First, let's calculate the mass of the solution:

m = (100.0 mL + 100.0 mL) * (1.00 g/mL)
m = 200.0 g

Next, let's calculate the change in temperature:

ΔT = (29.8°C - 23.0°C)
ΔT = 6.8°C

Now, we can calculate the heat released:

q = 200.0 g * 4.18 J/g·°C * 6.8°C
q = 5719.2 J

To convert the heat from joules to kilojoules:

q = 5719.2 J / 1000
q = 5.72 kJ

Therefore, the heat released in this reaction is 5.72 kJ.