Please help I have no idea how to get this. I ended up with -94 ? A coffee cup c

Question

Please help I have no idea how to get this. I ended up with -94 ?

A coffee cup calorimeter contained 50.0 mL of a 1.33 M KOH solution at an initial temperature of 24.3 degree C. A student quickly add 35.0 mL of 1.42 M HN03 to the calorimeter. The maximum temperature reached during the neutralization reaction was 29.1 degree C. Determine the delta H_rxn for this neutralization reaction in kj/mol of HNO_3. (You will need to first find q_solution). assume specific heat of solution is the same as specific heat of water = 4.18J/g degree C assume density of both KOH and HNO_3 are the same as water = 1.00 g/mL

Explanation / Answer

Heat released by reaction is absorbed by solution
qsolution = m*c*delta T
mass of solution, m = density * toltal volume
= 1.00 g/mL * (50 mL + 35 mL)
= 85 g

qsolution = m*c*delta T
= 85 * 4.18*(29.1 - 24.3)
= 1705.44 J

number of moles of KOH = M*V = 1.33 M * 50 mL = 66.5 mmol
number of moles of HNO3 = M*V = 1.42 M * 35 mL = 49.7 mmol

moles of acid and base reacted = 49.7 mmol = 0.0497 mol

H = -qsolution/mol reacted
= - 1705.44 J / 0.0497 mol
= - 343147 J/mol
= - 34.3 KJ/mol

Answer: - 34.3 KJ/mol

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