How many grams of alcl3 are produced when 3 grams of al(oh)3 react completely with excess hcl?

The  grams  of  Alcl3  that   are  produced   when  3  grams  of  Al(OH)3  react  completely   with   excess Hcl is  calculated   as   follows

write  the reacting  equation
Al(Oh)3   +  3HCl = AlCl3  +  3H20

find  moles  of  Al (Oh)3  used
moles=  mass /molar  mass
=  3g/ 78  g/mol=  0.0385  moles

by  use  of  mole  ratio  between  Al(OH)3  to  AlCl3  which  is  1:1 this  implies  that  the  moles  of AlCl3  is  =  0.0385  moles

mass  of  Alcl3  =  moles  x  molar  mass

=   0.0385 mol x133 .5  g/mol  =  5.14   grams

The correct answer is b.hoped i you