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QUESTION 15 When 80.4 g Al and 70.4 g Cl2 react according to the following equation, how much AlCl3 can be made, assuming 100
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Answer #1

2 AL +30 2 Alcz Giver, mass of Al = 80.ug. cz - 70.4 g Mole of Al= mass of AL = 80.4 26.98 molecular mass = 2.9799 male MolesLimiting reagent in this reaction is Cl2 . because mol of Cl2 / its coifficient is less than that of Al .

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