How to solve this chemistry question?

Question

Q. Sodium hydroxide (4.00 g) and water (100.0 g) are both at 20.0 °C. When the sodium hydroxide is dissolved in the water using a coffee cup calorimeter, the solution’s temperature reached a maximum of 29.6 °C. The specific heat of the solution is 3.98 J∙g^-1 ∙ ^oC ^-1 . Use NaOH as the limiting reactant and calculate its molar enthalpy of solvation in kJ/mole.

Answer 1

q = mC∆T

q = heat = ?

m = mass of water = 100.0 g

C = specific heat = 3.98 J/gº

∆T = change in temperature = 29.6 - 20.0 = -9.6ºC (negative because heat was released = exothermic)

q = (100.0 g)(3.98 J/gº)(-9.6º)

q = -3821 J = ∆Hrxn

To get ∆H per mole NaOH, we must find moles of NaOH used in the reaction.

moles NaOH = 4.00 g NaOH x 1 mol/40 g = 0.1 moles

Molar ∆H solvation = -3821 J/0.1 moles = -38,210 J/mol = -38.2 kJ/mole