Lead (ll) nitrate and ammonium iodide react to form lead(lI) iodide and ammonium nitrate according to the following reaction:
Pb(NO₃)₂ (aq) + 2 NH₄I (aq) → PbI₂ (s) + 2 NH₄NO₃ (aq) 
How many mL of a 0.178 M ammonium iodide solution is required to react with 232 mL of a 0.355 M Lead (ll) nitrate solution? And how many moles of lead(lI) iodide are formed from this reaction?

• A. 463 mL, 0.463 mol
• B. 925 mL, 0.165 mol
• C. 463 mL, 0.231 mol
• D. 925 mL, 0.0824 mol

Answer: (D)