Hope someone could fill up the rest of the blanks (plz show all the calculations
ID: 510831 • Letter: H
Question
Hope someone could fill up the rest of the blanks (plz show all the calculations) HCI (aq) NaoH (aq) HNO3(aq)+ NaOH (aq) 1. Volume of acid (mL 2. Molarity of acid (molL) M. 3. Initial temperature of acid, Tinitial (acid)(oc) 4. Volume of NaoH (mL 5. Initial temperature of NaoH, T nitial (NaOH) (OC) 6. Molarity of NaoH solution (moln) 7. Maximum temperature from graph, Tinal (oc) 3D e 2a 8. Average initial temperature of acid and NaOH (OC) 9. Temperature change of solution, ATsolution(oCO 50 mL 50 10. Volume of solution (acid plus base) (ml 11. Mass of solution, msolution (g) (assume the density of the solution is 1.00 g/mL 4.184 Jlg.oC 4184 Jlg.oC 12. Specific heat of solution 13. Heat absorbed by solution, qsolution (J) (show your calculation) 14. Heat absorbed by solution, qsolution (kJ) 15. Moles of H20 formed in the reaction (mo) (show your calculation) 16. Ahn (kumol H20) (show your calculation) 17. Average AHn (kulmo H20)Explanation / Answer
Ans. #1. Calculation of heat of neutralization: Trial 1:
Moles of HCl used = Molarity x Volume in liters
= (1.1 M) x 0.050 L
= 0.055 mol
Moles of NaOH used = Molarity x Volume in liters
= (1.1268 M) x 0.050 L
= 0.05634 mol
In a balanced stoichiometry reaction, 1 mol HCl is neutralized by 1 mol NaOH.
In the given reaction mixture, HCl is the limiting reactant.
# Total mass of the reaction mixture = 50 mL HCl + 50 mL NaOH
= 100 mL
= 100 g ; [assuming density to be 1.0 g/ mL]
#13/ 14. Heat absorbed by solution, q = = m c dT
= (100 g) x (4.184 J g-1 0C-1) x (30 – 22.7)0C
= 3054.32 J
= 3.054 kJ
#15. The neutralization of 1 mol HCl produces 1 mol H2O, the moles of water produced = 0.055 mol.
#16. Now, molar enthalpy of neutralization =
heat released per mol H2O formed
= - (3054.32 J) / 0.055 mol
= - 55533.09 J/ mol
= - 55.53 kJ/ mol
Thus, dHn = - 55.53 kJ/ mol H2O
Note: the -ve sign indicates that heating is being released during neutralization.
Ans. Part B. Calculation of heat of neutralization: HNO3 - NaOH
Since all variable, except temperature change, remains the same as above, many of the calculations will be common to both parts.
#13/ 14. Heat absorbed by solution, q = = m c dT
= (100 g) x (4.184 J g-1 0C-1) x (29.6 – 22.7)0C
= 2886.96
= 2.88696 kJ
#15. The neutralization of 1 mol HNO3 produces 1 mol H2O, the moles of water produced = 0.055 mol.
#16. Now, molar enthalpy of neutralization =
heat released per mol H2O formed
= - (2.88696 kJ) / 0.055 mol
= - 52.49 kJ/ mol
Thus, dHn = - 52.49 kJ/ mol H2O
#17. Average dHn = (dHn for HCl + dHn for HNO3) / 2
= [- 55.53 kJ/ mol H2O + (- 52.49 kJ/ mol H2O)] / 2
= - 54.01 kJ/ mol H2O
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