Question

You want to determine AH for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose AH is known: NaOH(aq) + HCl(aq) + NaCl(aq) + H2O(1) -57.32 kJ (a) Calculate the heat capacity of the calorimeter from these data: Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH Initial T of both solutions: 16.9°C Maximum T recorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL c of 1.00 M NaCl(aq): 3.93 J/gK L KJPC

Answer 1

density of 1.om Nall solution (d=1.04 g/me - Total volume (U) = loome . mass of solution (m) =du change 1 = 100g in to Temperature = 30.1-1609 Heat capacity of Nael soln (s) = 3.93319k - 13:2 0 13-2K moles of Naoh = mxu = 2.0 und /LX 0.050 - Oolo mol According to run 0.10 ud Nooh when react with 0-10 mil Hd, Heat produced = 5.732kJ 1 This heat is taken by nal solo and calorimeter. WIE

q water = mSAT S= 3.93J/g.k csalt) . = 3.935/9°) = 104943.935/g% *(304_16690°c 25511:72 J =5.51712 ku 9 cal i Ceal AT =Ceal (13.5%) 5.732 ku = q salt water + Veal 5.51712 t Ceal x 13.5 = Ceal *13.5°C = 0.21428KJ Heat Capacity fcalorinate feal ) = 0.0158 KJ loe