3) You would like to measure the heat of neutralization of an acid with a base....

the balanced reaction

HNO3(aq) + LiOH(aq) = H2O(l) + LiNO3(aq)

Part a

Mass of solution = volume x density

= (225 mL x 2) x 1 g/mL = 450 g

Heat absorbed by the solution

qsoln = mass x Cp x (T2-T1)

= 450 g x 4.184 J/gC x (22.3 - 18.1)C

= 7907.76 J

Part b

Heat absorbed by calorimeter

q cup = C cup x (T2-T1)

= 47 J/C x (22.3 - 18.1)C

= 197.4 J

Part c

heat released by reaction

qrxn = qsoln + q cup

= 7907.76 + 197.4

= 8105.16 J

Part d

Moles of HNO3 = molarity x volume

= 0.65 mol x 225 mL x 1L/1000 mL

= 0.14625 mol

Moles of LiOH = = 0.65 mol x 225 mL x 1L/1000 mL

= 0.14625 mol

ΔH = (qrxn / mol)

= 8105.16 J / 0.14625 mol

= 55419.90 J/mol x 1kJ/1000 J

= 55.42 kJ/mol

Part e
The balanced reaction

HNO3(aq) + LiOH(aq) = H2O(l) + LiNO3(aq)

Total ionic equation

H+(aq) + NO3-(aq) + Li+(aq) + OH-(aq) = H2O(l) + Li+(aq) + NO3-(aq)

Net ionic equation

H+(aq) + OH-(aq) = H2O(l)