Question 35 of 45 Sapling Learning A coffee cup calorimeter with a heat capacity

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Question 35 of 45 Sapling Learning A coffee cup calorimeter with a heat capacity of 4.40 J/°C was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.370 M AgNO3 was mixed with 50.0 mL of 0.530 M KCI. After mixing, the temperature was observed to increase by 2.90 . Calculate the enthalpy of reaction, tan, per mole of precipitate formed (AgCl). Assume the specific heat of the product solution is 4.15 J/(g·°C) and that the density of both the reactant solutions is 1.00 g/mL Calculate the theoretical moles of precipitate formed from AgNO3 (left) and KCI (right). Number Number moles D moles Number Calculate the heat change experienced by the calorimeter contents, Qcontents Number Calculate the heat change experienced by the calorimeter, cal Number Calculate the heat change produced by the solution process, Qsolution Using the mole values calcuated above, calulate AHsolution for one mole of precipitate formed. Number kJ/ mole O Previous 8 Give Up & View Solution O Check Answer NextExit Hint

Explanation / Answer

AgNO3(aq) + KCl(aq) ----> AgCl(s) + KNO3(aq)

no of mol of AgNO3 = 50*0.37/1000 = 0.0185 mol

no of mol of AgCl formed from AgNo3 = 0.0185 mol

no of mol of KCl = 50*0.53/1000 = 0.0265 mol

no of mol of AgCl formed from KCl = 0.0265 mol

heat change experienced by the calorimeter contents = m*s*DT

    = 100*4.15*2.9

    = 1203 j

heat change experienced by the calorimeter = c*DT

             = 4.4*2.9

            = 12.76 j

heat change produced = 1203+12.76 = 1215.76 joule

     limiting reactant = AgNo3

DHsol = -q/n

       = -1.2158/0.0185

       = -65.72 kj/mol

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