Moles of aluminum produced = 0.588
Further explanation
Given
30 g of Al2O3
Reaction
2Al203 ⇒ 4Al + 302
Required
moles of aluminum
Solution
mol Al2O3 :
= mass : MW
= 30 : 101,96 g/mol
= 0.294
From the equation, mol Aluminum :
= 4/2 x mol Al2O3
= 4/2 x 0.294
= 0.588
