Final answer:
The mass of silver nitrate required for the precipitation of chloride ions in a solution, we need to use stoichiometry and the balanced equation for the reaction. Given that 3.06 grams of solid AgCl are recovered from the reaction mixture, the mass of AgNO3 present in the reactants is 3.77 grams.
Step-by-step explanation:
To determine the mass of silver nitrate required for the precipitation of chloride ions in a solution, we need to use stoichiometry and the balanced equation for the reaction. From the balanced equation, we can see that the stoichiometric ratio between AgNO3 and AgCl is 1:1.
This means that 1 mole of AgNO3 is required to react with 1 mole of AgCl. To calculate the mass of AgNO3, we need to convert the mass of AgCl to moles, and then use the molar mass of AgNO3 to convert back to mass.
Given that the molar mass of AgCl is 143.32 g/mol, we can convert the mass of AgCl to moles by dividing it by its molar mass:
Moles of AgCl = 3.06 g / 143.32 g/mol = 0.021 moles
Since the stoichiometric ratio between AgNO3 and AgCl is 1:1, the moles of AgNO3 required would also be 0.021 moles. Finally, we can convert moles of AgNO3 to mass by multiplying it by the molar mass of AgNO3:
Mass of AgNO3 = 0.021 moles * 169.88 g/mol = 3.77 grams