If 0.683 grams of silver chloride is produced how much (mass) silver nitrate would need to be reacted

  The (mass)  silver  nitrate  would  needed  to  be  reacted    is  calculated   as  follow

   by  assuming   that    AgNo3  reacted  with  Bacl

that  is   2AgNo3  +  BaCl2  --->  2Agcl  +  Ba(No3)2 

find  the  moles   AgNo3  = 0.683g/169.87  g/mol  =4.02 x10^-3  moles

by  use  of  mole   ratio    between  Agcl  and  Agno3 which  is   2:2  this  implies  that   the  moles of  AgNO3 also  4.02  x10^-3

mass=  moles   x  molar   mass

=  (4.02  x10^-3)   x 143.37=  0.576  grams

The answer is a i think