What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

 " 13.5 g Al " ;

                    →  that is:  "13.5 grams of aluminum."

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Explanation:

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Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

 The reactants:  "aluminum (Al) " ;  and "chlorine (Cl) " ;  and:

 The product:    "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

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Note: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

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So:  We call solve using a technique known as:  "dimensional analysis" :

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  0.500 mol AlCl₃ *

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Note:  The units of "mol AlCl₃" cancel out to "1' ; and:

          The  units of "mol Al" cancel out to "1" ; and we are left with:

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 "   g Al ["grams of aluminum"] ;

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Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

 →  (0.500 * 26.98) g Al ;

    = 13.49 g Al ;

         →  Round to 3 (Three) significant figures;

         →  Since:  "0.500" has 3 (Three) significant figures:

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   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

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 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "" community!

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answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error