54g Al and 320g of oxygen gas react to form Al2o3 , calculate mass of Al2o3 formed and identify limiting reagent
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Aluminum reacts with oxygen in the following chemical reaction: Al + O2 → Al2O3. How many moles of Al2O3 are formed from the reaction of 6.38 mol O2 and 9.15 mol of Al?
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TUSHAR CHANDRA eNotes educator | CERTIFIED EDUCATOR
The chemical equation when aluminum reacts with oxygen is:
4Al + 3O2 --> 2Al2O3
3 moles of O2 react with 4 moles of Al to give 2 moles of Al2O3. The ratio of the number of moles of aluminum to the number of moles of oxygen required in the reaction should be 4/3.
When 6.38 moles of O2 and 9.15 moles of Al react, the ratio of aluminum to oxygen is 9.15/6.38 = 1.434. This shows that all of the oxygen is consumed before the aluminum and when the reaction ends the reactant aluminum is left.
As 3 moles of O2 give 2 moles of Al2O3, the number of moles of Al2O3 formed is 4.25 moles
The number of moles of Al2O3 formed is approximately 4.25 moles.