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14. For the reaction shown, find the limiting reactant for each of the initial amounts of reactants. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) a. ImolAl, 1mol02 b. 4molA1,2.6mol02 c. 16molAl,13mol02 d. 7.4 mol Al, 6.5 mol O2

Answer 1

Al is + 302 (9) 2A203 (5) ay I mol AS, Imol O2 to m, u mod sl react with 3 mal u2 then mol al react with 3 mol O₂ ne 0.75 mul o, but we have Imole Or te exces of O2 here, limiting reactant is Al. by - u mol Ae, 2.6 mol O2. 4 mol al react with 3 mol Or but we have only 2.6 mol O₂ Henne, limiting reactant is O2. CT 16 mod sl, 13 mol O₂. 16 mol 20 yeach with 3x H = V ote, 12 moi of O₂ but we have 13 mol Ole excess of oz. Mence, limiting reactant is Al. dy T4 mol se 6.5 mol O2.. 14 mol se react with 2 x 74 = 5.55 mol of a but we have bes mod Orie. excers of O₂. Hence, limiting reactant is Al.