Question

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

Answer 1

  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