Al2(SO3)3 + 6NaOH �¨ 3Na2SO3 + 2Al(OH)3

Question

If this reaction begins with 10 g Al2(SO3)3 and 10 g NaOH, what is the theoretical yield in grams Al(OH)3 and what is the limiting reactant?

Answer 1

Al2(SO4)3 + 6NaOH ==>3Na2SO4 + 2Al(OH)3
mols Al2(SO4)3 = grams/molar mass = ?
mols NaOH = grams/molar mass

Using the coefficients in the balanced equation, convert mols Al2(SO4)3 to mols Al(OH)3.
Do the same and convert mols NaOH to mols Al(OH)3.
It is likely that the two values will not agree which means one is not right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that number is the limiting reagent.

Using the smaller number, convert to grams. grams Al(OH)3 = mols Al(OH)3 x molar mass Al(OH)3. That is the theoretical yield .

Answer 2

3.44