Question

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

Answer 1

Answer:  " 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 * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

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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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 " $\frac{(0.500 * 2 * 26.98)}{2}$   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 "Brainly" community!

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