When 1.0 L solution Ba (NO3) 2 2.0 M at 25oC is mixed with 1.0 L solution of 2.0 M Na2SO4 in a calorimeter, the reaction that occurs to produce solid barium sulfate and temperature of the mixture increased to 28.1 oC. Calculate the enthalpy change per mole for the reaction that occurs. Assume that the calorimeter does not absorb the heat generated, the heat capacity of the solution is 4:18 J oC-1 g-1 and the solution density of 1.0 g mL-1

To calculate the enthalpy change per mole for the reaction, we can use the equation:

q = m × C × ΔT

Where:
q = heat absorbed or released by the solution (in Joules)
m = mass of the solution (in grams)
C = heat capacity of the solution (in J oC-1 g-1)
ΔT = change in temperature of the solution (in oC)

First, we need to determine the mass of the solution:
Since the density of the solution is given as 1.0 g mL-1, and the volume of the solution is 1.0 L,
We can convert the volume of the solution to grams using the density:
mass = volume × density
mass = 1.0 L × 1.0 g mL-1 × 1000 mL L-1 = 1000 grams

Next, we can calculate the heat absorbed or released by the solution, q:
q = m × C × ΔT
q = 1000 g × 4.18 J oC-1 g-1 × (28.1 oC - 25 oC)
q = 1000 g × 4.18 J oC-1 g-1 × 3.1 oC
q = 12998 J

Assuming that the reaction occurs between 1 mole of Ba(NO3)2 and 1 mole of Na2SO4, and that the reaction is the limiting reaction, the enthalpy change per mole for the reaction can be calculated as follows:

Enthalpy change per mole = q / number of moles of the limiting reactant

The number of moles of the limiting reactant can be calculated using the given molarity and volume:
number of moles = molarity × volume

For Ba(NO3)2:
number of moles of Ba(NO3)2 = 2.0 M × 1.0 L = 2.0 moles

For Na2SO4:
number of moles of Na2SO4 = 2.0 M × 1.0 L = 2.0 moles

Since both reactants have the same number of moles, 2.0 moles, we can choose either of them to calculate the enthalpy change per mole:

Enthalpy change per mole = q / 2.0 moles
Enthalpy change per mole = 12998 J / 2.0 moles
Enthalpy change per mole = 6499 J/mol

Therefore, the enthalpy change per mole for the reaction is 6499 J/mol.