Question

Given 25.0 g of Al, how many grams of the product Al2O3 could be produced?

Answer 1

4 Al + 3 O₂ = 2 Al₂O₃

or,   Al + (3/4) O₂ = (2/4) Al₂O₃

25 gm Al = (25/27) = 0.93 moles of Al

1 mole Aluminum produces 0.5 mole of  Al₂O₃

0.93 moles of Aluminum produces (0.5 X 0.93) = 0.465 moles of Al₂O₃

Molar weight of Al₂O₃ = ( 27X2) + (16 X3) = 102

0.465 moles of Al₂O₃ = 0.465 X 102 = 47.43 gms of  Al₂O₃