Question

second run (70.0 mL water) third run (50.0 mL water) 200g 1509 Zee first run (100.0 ml water) mass of CaCl2 (9) 20.09 mass of water (9) 2eeg mass of final soln (9) 2209 Ti, initial temperature of water (°C) 200 Ts, final temperature of 35.92°C water (°C) AT (°C) 15.920 heat absorbed by the 146605 solution (J) heat produced by the reaction (1) | 13.335 AHan (Jg CaCl2) 170g 20°C 4.60°C 20.6°C 14660 T 12.931 20.09 leeg 120g 20°c 49.18°C 29.18°C 14660 12.215 AHän (kJ molt CaCl2) mass CaCl2 mass water

how do I find the delta Hrxn based off of the information recorded?

Answer 1

Heat produced - . wt of cael o Harang (97 cade) = Frist runt 3- 13:33 zog qeach = -0.6665 5 97 DH is -ve because ryn is exothermie Ahnan = 1. - 12.93 ] -0.6965 J 97 2 rum 1. 209 . DH ram = -12-21 I - -0.6105 J 97 209 . . 3rd Run q cael . no. q moles = wt q cael Miest of caeli - 208 110.9 gmort = 0.18 mol of Caela

-13.33] ist rum! Dhwan 0.19 ma I -72.22 Jmolt = -72.22 kJ mort 1000 = -0.072 kJmont 1 and rim: Dhan -12.931 kJ molt 1000 x 0.18 =-0.0718 kJmol? -12.21 3rd rums!- O Hrana - ki molt 1000 * 0.18 -0.068 kJ molt grd Rum pstrum 2nd rom mass cael 209 mass water 2009 209 1509 209 1009 -0.2 a Oil 0.133
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