Stoichiometry - 37.8 grams of Aluminum are available for reaction with excess oxygen. Calculate the mass of the aluminum oxide that is produced?

1 Answer
Hriman
Mar 14, 2018

#71.4 "g Al"_2"O"_3#

Explanation:

  1. Write a balanced equation
    #4Al(s)+3O_2(g)->2Al_2O_3(s)#

  2. Determine Limiting Reagant
    Well, no calculations here because aluminum is reacted with EXCESS oxygen, so Al is limiting

  3. Stoichiometry
    #37.8 cancel("g Al")* (1cancel("mol Al"))/ (26.98 cancel("g Al")) * (2 cancel("mol Al"_2"O"_3))/ (4 cancel("mol Al"))* (101.96 "g Al"_2"O"_3)/ (1 cancel("mol Al"_2"O"_3))=#

#71.4 "g Al"_2"O"_3#

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