What mass of kcl is needed to precipitate the silver ions from 15.0 ml of 0.200 m agno3 solution?

This type of reaction occurs in the method of argentometric titration. This methods involves a silver solution that displaces to bond with the chloride ions in the sample to yield the precipitate, Silver Chloride (AgCl). The reaction between Silver Nitrate and the Potassium Chloride yields Silver Chloride and Potassium Nitrate:

KCl + AgNO₃ → AgCl + KNO₃

The balanced equation shows 1:1 ratio for all substances. These stoichiometric coefficients are useful in our calculations. Let's compute the moles of AgNO₃:

0.200 mol  AgNO₃/L * (0.015 L) = 0.003 moles of AgNO₃

Since 1 mole of AgNO₃ is equivalent to 1 mole of KCl, then, it would also need 0.003 moles of KCl. Knowing that the molar mass of KCl is 74.55 g/mol, the mass would be

0.003 moles * 74.55 g/mol = 0.224 g of KCl

I'm not sure but i will try to find out