if 20.0g of aluminum is reacted with 30.0g of chlorine to make 34.3g of aluminum chloride, what is the percent of yield of this reaction?

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  • 4 weeks ago
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    2 Al + 3 Cl2 → 2 AlCl3

    (20.0 g Al) / (26.98154 g Al/mol) = 0.741248 mol Al

    (30.0 g Cl2) / (70.9064 g Cl2/mol) = 0.423093 mol Cl2

    0.423093 mole of Cl2 would react completely with 0.423093 x (2/3) = 0.282062 mole of Al, but there is more Al present than that, so Al is in excess and Cl2 is the limiting reactant.

    (0.423093 mol Cl2) x (2 mol AlCl3 / 3 mol Cl2) x (133.3405 g AlCl3/mol) =

    37.610 g AlCl3 in theory

    (34.3 g actual) / (37.610 g theoretical) = 0.91199 = 91.2% yield AlCl3

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