Taking into account the stoichiometry of the reaction, 182.4 grams of O₂ must react to form 3.80 mol of Al₂O₃.
The balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of each compound is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Then it is possible to apply the following rule of three: If by reaction stoichiometry 2 moles of Al₂O₃ are formed from 96 grams of O₂, 3.80 moles of Al₂O₃ are formed from how much mass of O₂?
[tex]mass of O_{2}=\frac{3.80 moles of Al_{2} O_{3} x96 grams of O_{2} }{2 moles of Al_{2} O_{3} }[/tex]
mass of O₂= 182.4 grams
In summary, 182.4 grams of O₂ must react to form 3.80 mol of Al₂O₃.
Learn more:
