6.5 moles alcl3 reacts with 57.0g of naoh. how many grams of al(oh)3 will be produced

The equation for the reaction between NaOH and AlCl₃ is as follows;
3NaOH + AlCl₃ ---> 3NaCl + Al(OH)₃
the stoichiometry of NaOH : AlCl₃ is 3:1
3 moles of NaOH reacts with 1 mol of AlCl₃ to produce 1 mol of Al(OH)₃
the number of AlCl₃ moles reacted - 6.5 mol
molar mass of NaOH -(23 +16 +1) = 40 g/mol
the number of NaOH moles reacted = 57.0 g / 40 g/mol
 NaOH moles = 1.425 mol
either NaOH or AlCl₃ is in excess and other is the limiting reactant.
limiting reactant is the reactant whose number of moles are fully consumed during the reaction. the reactant that is in excess will have leftover moles that are remaining after the reaction.
If AlCl₃ is the limiting reactant, number of NaOH moles would be thrice the amount of AlCl₃ present,
then number of NaOH moles that should be present - 6.5 * 3 = 19.5 mol
however there are only 1.425 mol of NaOH present, therefore AlCl₃ is in excess.
Then NaOH is the limiting reactant,
the amount of products formed depends on the amount of the limiting reactant present.
stoichiometry of NaOH : Al(OH)₃ is 3:1
the number of Al(OH)₃ moles produced = number of NaOH moles reacted / 3
 number of Al(OH)₃ moles are - 1.425 mol /3  = 0.475 mol
molar mass of Al(OH)₃ = (27 +3*16 + 3*1) = 78 g/mol
mass of Al(OH)₃ produced = 78 g/mol * 0.475 mol = 37.05 g

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