If 10.0g of Al2 (SO3) is reacted with 10.0g of NaOH, determine the limiting reagent

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Chemistry

Sylvia Duran

25 March, 11:41

If 10.0g of Al2 (SO3) is reacted with 10.0g of NaOH, determine the limiting reagent

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Victoria Patrick · Answer 1

(10.0 g Al2 (SO3) 3) / (294.1544 g Al2 (SO3) 3/mol) = 0.033996 mol Al2 (SO3) 3 (10.0 g NaOH) / (39.99715 g NaOH/mol) = 0.25002 mol NaOH

0.033996 mole of Al2 (SO3) 3 would react completely with 0.033996 x (6/1) = 0.203976 mole of NaOH, but there is more NaOH present than that, so NaOH is in excess and Al2 (SO3) 3 is the limiting reactant.