Thermochemistry, can anyone help me with this calorimetry problem?

Question

In a constant pressure calorimeter, 199 mL of a 1.00 M alkaline earth metal (Me) chloride solution were added to 199 mL of a 1.00 M potassium sulphate solution. The total heat capacity of the calorimeter including all the contents was 1.75 kJ oC-1. The temperature in the calorimeter rose from 24.50oC to 28.42 oC.
Paying particular attention to sign and significant figures,
calculate the enthalpy change, in kJ, for the reaction:
Me2+ (aq) + SO42- (aq) = MeSO4 (s).
Me is the alkaline earth metal in question.
If you wish to use scientific notation, use the "e" format - eg. 7.31e4 = 73100 & 1.90e-2 = 0.0190.

Answer 1

The temperature increase = ΔT = 28.42ºC - 24.50ºC = 3.92ºC

Quantity of Heat absorbed by the calorimeter = Heat Capacity x Temperature increase

Quantity of Heat absorbed by the calorimeter = 1.75 x (-)3.92 kJ = -6.86 kJ (The -ve sign indicates that the reaction was exothermic and heat was lost to surroundings)

So, ΔH (Enthalpy Change) for the given mass of reactants = -6.68 kJ

Now, The above result was for 199 mL of 1.00 M of both reactants OR in other words, 0.199 L of 1 mole/litre i.e.=

0.199 moles of MeCl2 and 0.199 moles of K2SO4.

In the reaction-
0.199 mol of Me2+ and 0.199 mol of SO4 2- react to form 0.199 mol of MeSO4

Heat evolved when 0.199 mol of Me2+ reacts with 0.199 mol of SO4 2- = -6.68 kJ

So, heat evolved when 1 mol of Me2+ would react with 1 mol of SO4 2- = -6.68 kJ divided by 0.199 = -33.567 kJ mol-1

= -3.3567e1 kJ mol-1

Good Luck! I hope this helped.