Calculate the molar enthalpy of neutralization for the limiting reagent. The for

Question

Calculate the molar enthalpy of neutralization for the limiting reagent. The formula is; 2NaOH+H2SO4———> 2H2O+Na2SO4 Volume of NaOH is 50 ml for H2SO4 31 ml C is 1mol/l for both of them Initual Temperature is 20 Final temperature is 30 (same for both) (Assume the density of the solution is equal to that of water and the specific heat capacity is 4.18 J/g•Celsius) Calculate the molar enthalpy of neutralization for the limiting reagent. The formula is; 2NaOH+H2SO4———> 2H2O+Na2SO4 Volume of NaOH is 50 ml for H2SO4 31 ml C is 1mol/l for both of them Initual Temperature is 20 Final temperature is 30 (same for both) (Assume the density of the solution is equal to that of water and the specific heat capacity is 4.18 J/g•Celsius) The formula is; 2NaOH+H2SO4———> 2H2O+Na2SO4 Volume of NaOH is 50 ml for H2SO4 31 ml C is 1mol/l for both of them Initual Temperature is 20 Final temperature is 30 (same for both) (Assume the density of the solution is equal to that of water and the specific heat capacity is 4.18 J/g•Celsius)

Explanation / Answer

Qsoln= m*C*(Tf-Ti

Qsoln= (50+31)(4.184)(30-20) = 3389.04 J

assume

Qrxn = - Qsoln = - 3389.04

mol of NaOH is limiting, so mol of NaOH = MV = 1*50 = 50 mmol

HRxn = Qrxn/mol = -3389.04 * (50*10^-3) = -169.452 J/mmol = -169452 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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