In a styrofoam cup calorimeter, a 55ml sample of .6M HI at 18.46 degrees celsius is...

In a styrofoam cup calorimeter, a 55ml sample of .6M HI at 18.46 degrees celsius is allowed to mix with 86 ml of KOH at the same temperature. The final temperature of the reactoin mixture reached 21.96 degrees celsius. Calculate ∆H per mole of limiting reagent for this reaction. (Assume the density of the solution is 1 g/ml and the C_sp of water is 4.184 J/g)

Expert Solution

Total volume of solution = 55 + 86 = 141ml

mass of solution = volume * density

= 141*1 = 141g

q = mcT

= 141*4.184*(21.96-18.46)

= 2064.8J

HI + KOH -----------------> KI + H2O

no of moles of HI = molarity * volume in L

= 0.6*0.055 = 0.033 moles

HI + KOH -----------------> KI + H2O

heat energy released per mole = 2064.8/0.033 = 62569.7J/mole = 62.569KJ/mole

queen_honey_blossom answered 2 years ago

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