Step 1: Identifying the Complex
- The green complex formed is \[Co(NH_3)_6\]Cl\(_3\).
- This means that for each molecule of the complex, 3 chloride ions are available for precipitation.
Step 2: Moles of Complex and Cl\(^-\) Ions
- Given: 100 mL of 1M solution \( \Rightarrow \) Moles of complex = 0.1 moles.
- Each mole of the complex releases 3 moles of Cl\(^-\).
\[
{Moles of Cl\(^{-}\) = } 0.1 \times 3 = 0.3 { moles}
\]
Step 3: Reaction with AgNO\(_3\)
- AgNO\(_3\) reacts with Cl\(^-\) to form AgCl precipitate:
\[
Ag^+ + Cl^- \rightarrow AgCl {(s)}
\]
- Since each Ag\(^+\) reacts with one Cl\(^-\), the number of moles of AgCl precipitated is 0.3 moles.
Thus, the correct answer is 0.3 moles.