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Combustion Of Magnesium

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Determining the enthalpy of combustion of Magnesium

Om Vora
Ms. Mohoruk
Lab Partner: Ujjwal Nambiar
4/25/16
IB HL Chemistry 11

Purpose:

The purpose of this lab is to determine the molar enthalpy of combustion of Mg using Hess Law by manipulating three formulas, two of which will be found through the experiment and one value is already given.

Materials:

Refer to lab handout on the front of the page as no changes were made.

Procedure:

Refer to lab handout on the front of the page as no changes were made.

Qualitative results:
• Bubbles and vapor is released during the reaction of both Mg and MgO.
• Solutions in the calorimeter got really hot after the mixing the two. Sign of exothermic reaction.
• Small residue of …show more content…
It was determined by using values calculated by reacting Magnesium metal with HCl ( and Magnesium Oxide with HCl (. I first had to find the heat transferred (Q = mcΔT). All the values were either given or written down during the lab except for one, ΔT. The value of ΔT had to be extrapolated using an excel graph. Once the heat transferred between the Mg and HCl was determined (Q), I used the negative value to show the enthalpy change of the reaction that took place. I then had to convert the mass of the magnesium ribbon and magnesium oxide to moles and then divided the enthalpy change of the reactions that took place by the moles of the reactants. After finding the values for both Mg solid and MgO, I used the principles of Hess Law and found the enthalpy of combustion of Magnesium. The theoretical value is -601.7 , according to Dartmouth College, a private Ivy League research university. The value of the enthalpy of combustion found in this lab was . The percent error was . A percent error of 9.48% shows that the value is quite valid but its reliability could only be determined by having more trial done. A reason for the small difference in theoretical value and experimental value could be because of the kind of calorimeter used in this lab, a Styrofoam cup. Although it was a closed system, some heat could have been released the moment the reaction occurred, causing the value to be less,

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