Question

part B

Calorimetry PART A. Determination of the Calorimeter Constant For calculations, assume Density of HCI 1.0 g/mL of soin 4. 1841 C) "Cold water Hot water 509 121. SDS Mass Initial Temperature, Final Temperature, Change in temp. C 65.ooc DOC 11.12 .. .MA. . .. MAT. 9.-C.A. (show work in the box) Heat of hot water, w Heat of cold water, 5 313.685 3702.94J Heat of calorimeter, Cal 1610.84JJ C Part B. Enthalpy for Mg 8) + 2 HCl(aq) + MgCl2 fag) + H2 (2) Mass of HCI, Mass of Mg(s). Initial Temperature, Final Temperature, 15.0ML 0.24689 20.4°C 30.910 -9 -90 +9cal 9= m ., A7 9car = CAT (show work in the box) Heat of soln, coln Heat of calorimeter, cal Heat of rxn, 4x mol Mg, n AH, ME

Answer 1

1 mole Mg react with 2 mole HCl to form 1 mole MgCl2. that means 24.3040 gm Mg gives 95.211 gm MgCl2, so from 0.2468 gm Mg gives {(0.2468x 95.211)/24.3040} 0.967 gm.

m (soln) =0.967 gm, C (soln)= 4.184 J/ g oC, temperature difference = (30.917-20.4)= 10.517, Heat of the solution = 0.967x 4.184x 10.517 =42.55 J