Achemist has 4.0 g of silver nitrate and needs to prepare 2.0 l of a 0.010 m solution. how many grams of solid silver nitrate will be left over after the solution is made?

Molarity of a solution is defined as the number of moles of solute dissolved in 1 L solvent.
The molarity required in the solution to be prepared - 0.010 M
the volume of the solution to be prepared  - 2.0 L
In 1 L of solution there should be 0.010 mol
Therefore in 2.0 L solution, number of moles required - 0.010 mol/L x 2.0 L
the number of silver nitrate moles required - 0.020 mol
Molar mass of silver nitrate - AgNO₃ - 169.9 g/mol
Mass of silver nitrate needed to make solution - 0.020 mol x 169.9 g/mol 
Mass required - 3.4 g
chemist had 4.0 g and 3.4 g needed to make the solution
leftover solid AgNO₃ after the reaction is 4.0 - 3.4 = 0.6 g of AgNO₃

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !