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.

Related questions
Impact of this question
1002 views around the world
You can reuse this answer
Creative Commons License