Question

C. ΔH for Reaction of HCl_(aq) and NaOH_(s):
Volume of HCl: 0.055 L
Volume of water: 0.045 L
Volume total: 0.1 L
Molarity of HCl: 2.08 M
Mass of NaOH: 4 g
ΔT for reaction C: 23°C

number of moles of NaOH : 0.100 moles

molarity of the resulting sodium chloride solution : 1.00 M

ΔT : 23.0 °C

Calculate the value (calories) for the heat of reaction.

Calculate the ΔH for this reaction.

Answer 1

total volume of solution = 0.1L = 100ml

mass of solution = volume * density

                           = 100*1 = 100g

ΔT : 23.0 °C

q = mc$\Delta$T

     = 100*1*23

       = 2300cal

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

MOLARITY of NaCl = 0.1 moles

total volume of solution = 0.1L

no of moles of NaCl = molarity * volume in L

                              = 0.1*0.1 = 0.01moles

q = 2300Cal

    = 2300*4.184J

   = 9623.2J

$\Delta$H = q = 9623.2J

                  = 9623.2/0.01

                   = 962320J/mole

                    = 962.32KJ/mole >>>>answer