From the stoichiometry of the reaction, 0.74 moles of aluminium oxide is produced.
The equation of the reaction is;
4Al + 3O2 --> 2Al2O3
Number of moles of Al = 40.0/27 g/mol = 1.48 moles
Number of moles of O2 = 19.0 g/32 g/mol = 0.59 moles
Now;
4 mols of Al reacts with 3 moles of O2
1.48 moles of Al reacts with 1.48 moles × 3 moles/4 mols
= 1.11 moles
We can see that O2 is the reactant in excess.
Hence;
4 moles of Al produces 2 moles of aluminium oxide
1.48 moles of Al produces 1.48 moles × 2 moles/4 moles = 0.74 moles of aluminium oxide
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