Question

Solid aluminum (Al) and oxygen (0) gas react to form solid aluminum oxide (A1,0z). Suppose you have 9.0 mol of Al and 11.0 mol of O, in a reactor Calculate the largest amount of Al,0, that could be produced. Round your answer to the nearest 0.1 mol. mol P

Answer 1

First we have to find the limit reagent (out of Al and 0 ) Limiting seagent is a substance ( Reactants) which is limited amount of reaction and much product can be formed present in decides how y mo? 3 mol Balanced gtm b/w All and O, is: AAL (S) + 0, ) < 2A03 () From the reaction gt is clear that: - 4 mol of Al aequiser 3 mol of O or I mor of Al requiser 3 mol of Geren moles of Al = 9 mol 9 mot of Al will require = 2 x mol of Oz = 6.75 mol Requised moles of O = 6.75 mol - We have (Given) a ll mol Therefore 0 is passent in excess and is called excess reagent and the other reactant, Al is paesent in limited - amant and is called Limiting reagent. so Aluminium decides how much Aloz can be foomed.

C - coefficient Trick to find Limiting Reagent : Gse foomula no. of moles Stoichiometric coefficient (5.c) The Substance that gives the smallest number is the - limiting reagent. * write the Balanced eque 4 AD (8) + 30, (3) ► 2Al2Oz (5) 4 Stoichiometric 3 sie - no. of mol For = For Al = nostre portes - 7 = 2.25V Al 2.25 ✓ SIC For Of = normal moes - 3.66 v since Al gives the smallest number. . Al is the limiting reagent. To find amount of Al₂O3 4 Al +30 2 Al₂O3 From the reaction of is clear that - 4 mol of Al reacted to give 2 mol of Alle or I mol of Al reacted to give 22 mol of Aloe Given moles of Ae = 9 mol 9 mot of Al will react to give = 2 x mol of Aboz = 4,5 mol of Aloe Hound off to 0.45x10' r Ans is 0.45X 10' mob - or you simply write - - 0.4