Question

Calculations PART I-Mg 1. Calculate the heat energy. q. released by the reaction of Mg and HCl using the combined mass of all the reactants. 2. Calculate the enthalpy of this reaction, AH , in kJ/gram using the grams of Mg. (Note the sign of AH in your answer). 3. Calculate the enthalpy of this reaction, AHran, in kJ/mol using the molar mass of Mg. (Note the sign of AH in your answer). PART II - MgO 4. Calculate the heat energy. q. released by the reaction of MgO and HCl using the combined mass of all the reactants. 5. Calculate the enthalpy of this reaction, AH re, in kJ/gram using the grams of MgO. (Note the sign of AHrun in your answer). 6. Calculate the enthalpy of this reaction, AHrxa, in kJ/mol using the molar mass of Mgo. (Note the sign of AHre in your answer). More Calculations 7. Use Hess' Law to demonstrate how the three equations listed in the introduction are used to determine the heat of formation of Mgo. Then calculate the heat of formation of Mgo. 8. The accepted value for (Egn. 1) is -462 kJ/mol Mg. Calculate your % error. 9. The accepted value for (Egn.2) is -146 kJ/mol Mgo. Calculate your % error. 10. The accepted value for AH is -602 kJ/mol Mgo. Calculate your % error.

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Answer 1

I Given - My Reaction clata: Given mass of a meter of Mg Ribbon But 25 cm of Mg is used in reaction Hence mass of mg - Orig. Volume of Hcl used = 49.0 ml. - Yue = 49.mL. (19) AL = 49.g. The total mass of the reactants = 49. g my change in Temperature AT = 52.8-216 _ = 31.2°C We need to we specific heat of water C4.18 J/9.0c Now Heat Released q = m. c. AT 2 49.x 4:18 x 312 x= 6.513_kI

Here I fina > Tinitial Therefore a value is positive and energy is released so it is exothermice 2 Since the amount of energy left the system and enthalpy is negative AHL - =-6.513_KI SH=0 SHEN DHEXn = Ahlin where ne mase of Mg in grans = -6.5.13 kg/g SHARE - 65.13 +3/7]

3) Lo SELL -6,513 KI and sH - AHA Now, considering molar mass of my --------- the mass of Mg , - gm of mg /mol 0.1g 24.31 g/mol < 0.004113 -6.513. KJTY 0.004113 mol Xh. _=1583.5 kJ/mol Altersem = = 15.625x16 KT mod

4 Given vorherne of Hel. 510 mL a 510 .m - 5/8 eur change in temperaturu AT.- 2%=4.-21. Specific heat of water - 4. 183/gre Now Heat released q = m.c.st = 51x 4.18 X.6.07 = 14.28.3 ) 19 1:428 KT Here I final Tinitial & hence of is positive and energy -------www is released.

5) Since energy is released, enthalpy of system is negative DH=-q = -1,428_KJ . We know AH = D. DH Dhxo = AHO -1.428 kJ os q 6 (Henn 2-2.856 ksIgrana By considering molar volumast of Mgo - Molar man of mg + m.mofo _z_40.304 Now, n = 0.5g)| (40.304 g/ md) 00124_moles Attran = AH = -1.428 kJ 0.0124 md -DH AN = -115.16 kJ/mol -1.428 1st