Part A: Heat of Neutralization: HCI and NaoH Reaction HCI(aq) NaoH(ag) NaCl(aq)

Question

Part A: Heat of Neutralization: HCI and NaoH Reaction HCI(aq) NaoH(ag) NaCl(aq) H2000 Fill in the experimental d (with units and correct significant figures). Below each calculated value, show work and Original calculator answer, Use this original calculated value for consecutive calculations. mass of solution (Note assume solution density 1.01 g/mL) 2.470. Boy. 4.07 J/(g °C) Cal 85.0 JPC -0.9993 M Molarity of HCI 1.0 lass M Molarity of NaOH Moles of Limiting Reactant: as there a limiting reagent? Explain if yes moles of limiting reactant HCl or NaOH) Experimental kJ/mol AH moles of limiting reagent Calculate the standard enthalpy of neutralization, AHer using the net ionic reaction (Reaction D) for HCI (aq) and NaoH (aq) and the standard molar enthalpies of formation found in Appendix C in Ebbing. Assuming AH is the theoretical value and AH is the experimental value, calculate the percent enor. Meet briefly with students from yourlab bench and list four possible reasons belaw aslo why your percent emor for AH neu is not zero, Focus on apparatus, methods and assumptions, rather tban possible "blunders".

Explanation / Answer

part a

from the literature we can find out heat of formation values of NaOH,HCl, NaCl, H20

NaOH= -426.7 KJ/mol

HCl = -92.3 KJ/mol

NaCl = - 411 KJ/mol

H20 = 285.8 KJ/mol

so H nueralization = -411 - 285.8 + 92.3 + 426.7

= - 177.8 KJ/mol

mass of sol =1.01 g

so volume = 1ml

since molarity of NaOH is greater than HCl that means HCL is The limiting reactant.

moles of limitimg reactant = 0.5* 1.0125 = 0.4965 milli moles

From experimental values we can find that

qr= 2137.4 J

H nuetralization = 433.31 KJ

part b

besy value is = 456.975 KJ/mol

part c

best value is = 143.475 KJ/mol

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