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3 years ago

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

Chemistry

1 answer:

3241004551 [841]3 years ago

3 0

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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