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

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 =

putting the values in the equation

NaOH =

           = 21.1 MOLES OF NaOH

Al =

   = 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

=

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

the answer would be b i believe