Given #DeltaH_f^@# for #Al_2O_3#, how much aluminum is formed if #838*kJ# of energy are expended? #2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)# #DeltaH^@=-1676*kJ*mol^-1#

1 Answer
anor277
Aug 30, 2015

You have the equation; you also have the energy transferred during the reaction. Treat the energy required as a reactant.

Explanation:

#Al_2O_3(s) + 1676# #kJ rarr 2Al(s) + 3/2 O_2(g). #

You have the balanced stoichiometric equation. If you had produced 1 mol of aluminum, then by the reaction's stoichiometry you would know that you would require a 1/2 mol quantity of alumina as a reactant, and also #838# #kJ# energy. So treat the energy required as the limiting reagent. Please repost your answer here.

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