When 100 mL of Ba(NO3)2 solution at 25 degrees Celsius is mixed with 100 mL solution CaSO4 solution at 25 degrees Celsius in calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1 degrees Celsius. Assuming that the calorimeter absorbs only a negligible quantity of heat and the specific heat capacity of the solution is 4.184 J/g.degrees Celsius, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change of this reaction.

Ba(NO3)2 + CaSO4 ==> BaSO4 + Ca(NO3)2

mols Ba(NO3)2 = M x L = ?. No M given in the problem.

To calculate the enthalpy change of this reaction, we can use the equation:

q = m × c × ΔT

Where:
q is the heat transferred (in joules),
m is the mass of the solution (in grams),
c is the specific heat capacity of the solution (in J/g.degrees Celsius),
ΔT is the change in temperature (in degrees Celsius).

First, we need to find the mass of the solution:

100 mL of Ba(NO3)2 solution has a density of 1.0 g/mL, so the mass is 100 grams.

Next, we need to calculate the change in temperature:

ΔT = final temperature - initial temperature
ΔT = 28.1°C - 25°C
ΔT = 3.1°C

Now, we can substitute the values into the equation:

q = 100 g × 4.184 J/g.°C × 3.1°C
q = 1297.024 J

Since the reaction is exothermic (the temperature increased), the heat transferred (q) is equal to the enthalpy change (ΔH) of the reaction.

Therefore, the enthalpy change of this reaction is 1297.024 J (or -1297.024 J if you prefer to indicate the exothermic nature of the reaction using a negative sign).