Asked by Philip
Calculate the amount of heat released in the complete combustion of 8.17 grams of Al to from Al2O3 at 25 celcius and 1 atm. The delta enthalphy change in the heat of formation for Al2O3 = 1676 KJ/mol
4Al(s) + 3O2(g) --> 2Al2O3(s)
4Al(s) + 3O2(g) --> 2Al2O3(s)
Answers
Answered by
DrBob222
dH is 1676 kJ/mol Al2O3 and the equation has 2 mols Al2O3 in it.
2*1676 kJ x (8.17g Al/4*atomic mass Al) = ?
2*1676 kJ x (8.17g Al/4*atomic mass Al) = ?
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