If you use 1 mole of naoh, how much naal(oh)4 is produced

             81.97 g of NaAl(OH)₄

Solution:
              The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,

                 2 NaOH  +  2 Al  +  6 H₂O    →    2 NaAl(OH)₄  +  3 H₂

According to this equation,

          2 Moles of NaOH produces  =  163.94 g (2 mole) of NaAl(OH)₄
So,
       1 Mole of NaOH will produce  =  X g of NaAl(OH)₄

Solving for X,
                     X  =  (1 mol × 163.94 g) ÷ 2 mol

                     X  =  81.97 g of NaAl(OH)₄