Sapling Learning A coffee cup calorimeter with a heat capacity of 5.00 J/°C was

Question

Sapling Learning A coffee cup calorimeter with a heat capacity of 5.00 J/°C was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.340 M AgNOs was mixed with 50.0 mL of 0.520 M KI. After mixing, the temperature was observed to increase by 4.46 . Calculate the enthalpy of reaction, ton, per mole of precipitate formed (Ag),Assume the specificheat of the product solution is 4.17 J/(g. ) and that the density of both the reactant solutions is 1.00 g/mL Calculate the theoretical moles of precipitate formed from AgNOs (left) and KI (right). Number Number moles moles Number Calculate the heat change experienced by the calorimeter contents, dcontents Number Calculate the heat change experienced by the calorimeter, goD Number Calculate the heat change produced by the solution process, 9solution using the mole values calcuated above, cal late for one mole of precipitate formed. ok Number kJ/mole

Explanation / Answer

a)

theoretical moles:

mol of AgNO3 = MV = 50*0.34 = 17 mmol = 0.017 mol --> Forms 0.017 mol of AgI

mol of KI = MV = 50*0.52 = 26 mol = 0.026 mol --> forms 0.026 mol of AgI

overall --> only 0.017 mol of AgI forms

b)

find Qcontent

Qcontent= mwater*Cpwater*(Tf-Ti) = (50+50)*4.17 * (4.46) = 1859.82 J

For calorimter:

Qcal = Ccal * dT = 5 J/C * 4.46 = 22.3 J

Qsolution = 1859.82 +22.3 = 1882.12 J (absorbed)

then

HRxn = Qsoln / moles = -1882.12 / 0.017 = -110712.941 J/mol = -110.713 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.