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

It is a mixture. simply put, a compound must be pure, that is, of only one chemical structure and composition. a mixture is made of different chemical substances (i.e. elements, molecules, compounds) that are not chemically bonded to each other, but instead physically grouped together, as in the case of gasoline (a homogeneous solution).