Question

C. ΔH for Reaction of HCl_(aq) and NaOH_(s): CALCULATE DELTA H
Volume of HCl: 0.0550 L
Volume of water: 0.0450 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Mass of NaOH: 3.44 g
ΔT for reaction C: 17°C

Answer 1

Answer – There are given,

Volume of HCl: 0.0550 L
Volume of water: 0.0450 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Mass of NaOH: 3.44 g
ΔT for reaction C: 17°C

Total volume of solution = 0.100 L means 100 mL we assume the density of solution is 1.0 g/mL

So mass of solution = 100 g

Reaction - HCl + NaOH -----> H₂O + NaCl

We need to calculate the moles of of NaOH and HCl

Moles of HCl = 1.95 M * 0.0550 L = 0.107 moles

Moles of NaOH = 3.44 g / 40.0 g.mol^-1 = 0.086 moles

So limiting reactant is NaOH and total mass of solution = 3.44 g +100 g = 103.44 g

We know, q = -∆H

So, q = m*C*∆t

         = 103.44 g * 4.184 J/g^oC*17°C

         = 7357.5 J

So, solution absorb the 7357.5 J , means in reaction 7357.5 j heat gets released

So, ∆H = -7357.5 J

             = -7.357 kJ

∆H in kJ/mole = -7.357 kJ / 0.086 mole = -85.5 kJ/mol