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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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.
First We have to balance the given reaction
4Al + 3O2 ----> 2Al2O3
Moles of O2 = 6.38
Moles of Al = 9.15
Second find the limiting reactant.
Divide the given moles form Stoichiometric coefficient
For O2 = 6.38/3 = 2.126
For Al = 9.15/4 = 2.287
Since O2 has less value that is 2.126 than Al that is 2.287, O2 is the limiting reactant in this reaction.
Third compare the ratio of limiting reactant coefficient with the product coefficient.
That is the moles of O2 and Al2O3 that is 3:2 ratio.
Given moles of O2 is 6.38.
So we have to find how moles of Al2O3 is formed when 6.38 moles of O2 is reacting.
Since they are in the ratio of 3:2, 4.25 moles of Al2O3 will be formed when 6.38 moles of O2 is reacted.
Calculation ... (6.38 * 2)/3 = 4.25 moles of Al2O3.
Thus 4.25 moles of Al2O3 is formed when 6.38 moles of O2 and 9.15 moles of Al is reacting.
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