a coffee cup calorimeter is used to measure the heat of neutralization for this

Question

a coffee cup calorimeter is used to measure the heat of neutralization for this acid-base reactin. HBr + NaOH --> NaBr +H20. 35.00 mL of 0.225 M HBr are reacted with 50.0 mL of ).375 NaOh. Both solutions are initially at 20.500 deg C. After the reaction has finished, the temperature of the mixture is 21.646 deg C. The heat capacity of the calorimeter is 33.9 J/deg C and the specific heat of the mixture is 4.184 J/deg C. The density of the solution is 1.02 g/mL. 1. What is the total heat generated in experiment in J? 2. What is the number of moles of reaction? 3. what is the heat of neutralization for this reaction?

Explanation / Answer

You had

35 x 0.225 = 7.875 milli moles of HBr (moles = volume x concentration)

and 50*0.375 = 18.75milli moles of NaOH

so all the Hbr will react, butnot all the NaOH

you'll make 7.875 milli moles of water (that's answered part (a))

Then you use Q = mcdeltaT

so

Q = (85*1.02) x 4.184 x 1.146

= 415.7147088 joules or 0.4157147088 kiloJoules

then you divide the kilojoules by the moles to get the kJ per mole

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