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