How many aluminum atoms are in 3.28 g of al2s3?

m Al₂S₃: 27g×2 + 32g×3 = 150 g/mol

150 g/mol  Al₂S₃ ==>> 2 moles of Al ==>> 12,04·10²³ atoms of Al

150g 12,04·10²³  Al
3,28g X 
X = (3,28×12,04·10²³)/150g
X = 0,2633·10²³ = 2,633·10²²  aluminium atoms

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