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Which would be a source of more energy, the formation of 6 mol of magnesium oxide, MgO, or the formation of 2 mol of aluminum oxide, $\mathrm{Al}_{2} \mathrm{O}_{3} ?$ Explain.
Oxidation of $\mathrm{Mg}$ -$2 M g+O_{2} \rightarrow 2 M g O$For every one mole formation of MgO we need onemole of MgSo number of mole of Mg $=6$Heat of formation $=602 \mathrm{kJ} / \mathrm{mol}$Total energy $=6 \times 602=3612 \mathrm{kJ}$
Oxidation of Al-$4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3}$For every one mole formation of $A l_{2} O_{3}$ we need twomole of AlNumber $A l_{2} O_{3}$ mole $=2$ molSo number of mole of $\mathrm{Al}=2 / 2=1 \mathrm{mol}$Heat of formation $=838 \mathrm{kJ} / \mathrm{mol}$Total energy $=1 \times 838=838 \mathrm{kJ}$
Thus getting 6 mole Mg requires more energy.
Chemistry 102
Chemistry 101
Chapter 21
Controlling Energy
Section 2
Heat of Formation
Thermodynamics
Thermochemistry
Drexel University
Brown University
University of Toronto
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question eight deals with the Delta H of formation of two different oxides, magnesium oxide and aluminum oxide. And it wants to know which is going to produce more energy, the formation of six moles of magnesium oxide or two moles of aluminum oxide. So to form six moles of magnesium oxide, we need to look at the Delta H of formation for one mole of magnesium oxide on page 2 55. We see that for every one mole we're going to release 602 killer jewels or with six moles, will release so negative 3612 killer jewels for two moles of aluminum oxide. We need to look at the delta of formation of aluminum oxide for every one mole. It's negative 1676 killer jewels. So if we multiply that by two, we get negative. 3000 352 kill jewels. So it turns out we're going to get Mawr energy out of the formation of six moles of magnesium oxide than two moles of aluminum oxide
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