5. When 75.0 mL of 1.10 M HNO, was reacted with 75.0 mL of 1.25 M NaOH in a cons

Question

5. When 75.0 mL of 1.10 M HNO, was reacted with 75.0 mL of 1.25 M NaOH in a constant-pressure calorimeter, the temperature of the solution rose by 7.4 °C. a. Write a balanced complete equation for the reaction. b. Determine the moles of limiting reactant consumed, assuming the reaction goes to completion. (SHOW WORK!) c. Calculate the enthalpy of neutralization for this reaction in kJ /mole limiting reactant. (SHOW WORK!) Don't forget the sign on dH! Assume the density and specific heat of the solution are the same as for pure water. (See Question 1.) 7-8 Calorimetry: Enthalpies of Reactions ERIMENT 7 23-11

Explanation / Answer

a)

the reation

HNO3 + NaNOH = NaNO3 + H2O

b)

mol of HNO3 = MV = 1.1*75*10^-3 = 0.0825

mol of NaOH = MV = 1.25*75*10^-3 = 0.09375

clearly the acid is limiting

so

mol of acid = 0.0825

c.

apply

HRXN = -Qsolution/moles

Qsoln = m*C*(Tf-Ti) = (75+75)(4.184)(7.4) = 4644.24 J

HRxn = -4644.24 / (0.0825)

HRxn = --56293.81J/mol

HRxn = -56.29 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---