Table included: find Qwater, Qcal, Qrxn, moles reacted of the limiting reagent a

Question

Table included: find Qwater, Qcal, Qrxn, moles reacted of the limiting reagent and delta Hn. With questions.

Enthalpy of Neutralization

Concentration of NaOH

1.1 M

Initial temperature of NaOH

20*C

Volume of NaOH

45mL

Concentration of HCl

1.2 M

Initial temperature of HCl

19C

Volume of HCl

45mL

Final temperature of mixture

33*C

Average initial temperature (HCl and NaOH)

19.5*C

Temperature change, (delta T)

13.5*C

Mass of final mixture (assume density of mixture is 1.00g/mL)

90g

Specific heat capacity for H2O

4.18 J/*C

Qwater

Qcal

Qrxn

Moles reacted of the limiting reagent

DeltaH*n (Qrxn per mole reacted)

NaOH (s) + HCl (aq) NaCl (aq) + H2O (l)

Questions:

1. Show the required calculations to determine the calorimeter constant including heat released by the hot water and heat absorb by the cold water.

2. Show the required calculations to determine the enthalpy of neutralization (delta Hn) including the Qwater and Qcalorimeter.

3. Show the required calculations to determine the enthalpy of dissolution for NaOH including the Qwater and Qcalorimeter.

Please show calculations for all, including for the table.

Enthalpy of Neutralization

Concentration of NaOH

1.1 M

Initial temperature of NaOH

20*C

Volume of NaOH

45mL

Concentration of HCl

1.2 M

Initial temperature of HCl

19C

Volume of HCl

45mL

Final temperature of mixture

33*C

Average initial temperature (HCl and NaOH)

19.5*C

Temperature change, (delta T)

13.5*C

Mass of final mixture (assume density of mixture is 1.00g/mL)

90g

Specific heat capacity for H2O

4.18 J/*C

Qwater

Qcal

Qrxn

Moles reacted of the limiting reagent

DeltaH*n (Qrxn per mole reacted)

Explanation / Answer

Calculate the heat

m = 45 mL + 45 mLO = 90 mL = 90 g

Qwater = m*C*(Tf-Ti) = 90*4.184*(33-20) = Q = 4895.28 J

Qcalorimter = 4895.28

Q = 4895.28 J

moles reacted of limiting reactant = 1.1*45 mL = 49.5 mmol or 49.5*10^-3 mol

HRxn = -Q/mol = (-4895.28)/(1.1*45) = -98.89 kJ/mol

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