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Calorimetry Question

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billyboy:
Nitric acid reacts with potassium hydroxide as follows:

HNO3 (aq) + KOH(aq) --> KNO3 (aq) + H2O(l)

In an experiment a student placed 55.0 mL of 1.3 mol/L HNO3 and 55.0 mL of 1.3 mol/L KOH into a
calorimeter at 23.5 ̊C. The temperature rose to 31.8 ̊C after the reaction was complete. Calculate the heat
of reaction (∆HR) in kJ/mol of acid. Assume that the specific heat capacity and density of both solutions
are 4.2 J/g ̊C and 1 g/mL respectively.

I've been stuck on this question for a while, I just keep getting the wrong answer.

Borek:
Show how you got the wrong answer and we will start from there.

billyboy:
c= 4.2 j/g*C
deltaT = 31.8*C - 23.5*C
          = 8.3*C

mass of HNO3 = 55.0mL/1 x 1g/1mL
                      = 55.0g

Q= mcDeltaT
  = (55.0g)(4.2j/g*C)(8.3*C)
  = 1917 x 1kj/1000j
  = 1.917kj

n= m/M
  = 55.0g/63.02g/mol
  = 0.873mol

DeltaH= -Q/n
          = -(1.917kj)/0.873mol
          = -2.20kj/mol

But the answer should be -54kj/mol

Orcio_87:
@billyboy


--- Quote ---Q= mcDeltaT
  = (55.0g)(4.2j/g*C)(8.3*C)
  = 1917 x 1kj/1000j
  = 1.917kj
--- End quote ---

Volume (mass) of solution is greater than that!


--- Quote ---n= m/M
  = 55.0g/63.02g/mol
  = 0.873mol]
--- End quote ---

Concentration (and quantity) of HNO3 is given text!

billyboy:
I'm still confused because the mass of HNO3 in my mind can't be higher than 55.0g because it's a 1:1 ratio between mL and grams in this question.

Also with the concentration, meaning 1.3mol/L, I still get the wrong answer:

55.0mL x 1L/1000mL
= 0.055L

1.3mol x 0.055L
= 0.0715mol

DeltaH= -Q/n
          = - (1.917kj) / 0.0715mol
          = 26.81kj/mol

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