Question

Can someone please help me solve this? I've been stuck on this question for a long time and I'm still having trouble. Could you please also include the steps on how you would solve this, I have an exam coming up and somehow I still can't figure this out!

Question:

In a coffee-cup calorimeter, 1.74 g KOH is added to 125 ml of 0.80 M HCl. The following reaction occurs:

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)

Due to the reaction, the temperature of the solution changes from 39.0°C to 45.6°C. Given that the density of the HCl solution is 1.00 g/ml, and the specific heat of the final solution is 4.18 J/g·K, calculate the enthalpy change for this reaction in kJ/mol KOH.

________ kJ/mol KOH

Thank you!

Answer 1

no of moles of KOH   = W/G.M.Wt

                                  = 1.74/56   = 0.031moles

no of moles of HCl   = molarity * volume in L

                                  = 0.8*0.125   = 0.1moles

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)

The volume of solution = 125ml

The density of solution = 1g/ml

mass of solution   = volume * density

                             = 125*1 = 125g

q   = mcT

     = 125*4.18*(45.6-39)

      = 3448.5J

H = q = 3448.5J

H   = 3448.5J/0.031mole

          = 111242J/mole

          = 111.242KJ/mole of KOH >>>>>answer

Answer 2

no of moles of KOH   = W/G.M.Wt

                                  = 1.74/56   = 0.031moles

no of moles of HCl   = molarity * volume in L

                                  = 0.8*0.125   = 0.1moles

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)

The volume of solution = 125ml

The density of solution = 1g/ml

mass of solution   = volume * density

                             = 125*1 = 125g

q   = mcT

     = 125*4.18*(45.6-39)

      = 3448.5J

H = q = 3448.5J

H   = 3448.5J/0.031mole

          = 111242J/mole

          = 111.242KJ/mole of KOH >>>>>answer