Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HCl in a calorimeter. Both solutions were initially at 35.00°C and the final temperature of the resulting solution was recorded as 37.00°C. Write a balanced chemical reaction for the neutralization reaction between aqueous NaOH and HCl, determine the number of moles of water formed in this reaction, and calculate the overall heat change of the solution. Assume 1) that no heat is lost to the calorimeter and 2) that the density and the heat capacity of the resulting solution are the same as water (heat capacity = 4.184 J/g°C and density = 1.00 g/mL).

Answer 1

no of moles of NaOH = molarity * volume in L

                                   = 0.3*0.1 = 0.03moles

no of moles of HCl = molarity * volume in L

                                   = 0.3*0.1 = 0.03moles

total volume of solution = 100 + 100 = 200ml

mass of solution = volume * density

                            = 200*1 = 200g

          HCl(aq) + NaOH(aq) ----------> NaCl(aq) + H2O(l)

        1 mole of HCl react with 1 mole of NaOH to gives 1 mole of H2O

        0.03 moles of HCl react with 0.03 moles of NaOH to gives 0.03 moles of H2O

no of moles of H2O = 0.03 moles

q   = mc$\Delta$T

     = 200*4.184*(37-35)

      =1673.6J

q   = $\Delta$ H   = -1673.6J

$\Delta$H   = -1673.6J/0.03mole = -55786J/mole of H2O

                                                = -55.786KJ/mole of H2O >>>>answer