Question

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

Answer 1

`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`