Respuesta :

ardni313

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