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
