How many moles of aluminum will be produced from 30.0 g of Al2O3 in the following reaction? 2A1203 - 4A + 302

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

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