How many moles of Al2O3 are formed from the reaction of 6.38 mol O2 and 9.15 mol Al? With Explanation please !

Moles of Al₂O₃ formed : 1.42

Given

6.33 mol O₂

9.15 mol Al

Required

moles of Al₂O₃

Solution

Reaction

4Al + 3O₂ ⇒ 2Al₂O₃

Find limiting reactant (mol : coefficient) :

Al = 9.15 : 4 = 2.29

O = 6.38 : 3 = 2.13

O₂ as a limiting reactant(smaller ratio)

Mol Al₂O₃ based on mol O₂ as a limiting reactant

From equation, mol ratio Al : Al₂O₃ = 4 : 2, so mol Al₂O₃ :

= 2/3 x mol O₂

= 2/3 x 2.13

= 1.42

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

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