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ArbitrLikvidat [17]
3 years ago
10

If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically

form?
Chemistry
1 answer:
Phantasy [73]3 years ago
3 0

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

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Explanation:

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6 0
1 year ago
What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?
3241004551 [841]

Answer:  " 13.5 g Al " ;

                    →  that is:  "13.5 grams of aluminum."

<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: 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) " .

____________________________

The "balanced chemical equation" is:

____________________________

        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

_____________________________

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

____________________________

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

____________________________

  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

____________________________

<u>Note</u>:  The units of "mol AlCl₃" cancel out to "1' ; and:

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

____________________________

 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

____________________________

<u>Note</u>: 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:

____________________________

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

____________________________

 Hope this is helpful!  

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

____________________________

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