Answer to Question #305504 in General Chemistry for Mcquay

Question

Pb(NO₃)₂ + 2 NaI = PbI₂ + 2 NaNO₃

How many moles of lead (II) nitrate are needed to react completely with 10 moles of sodium iodide?

Answer 1

Pb(NO₃)₂ + 2NaI = PbI₂ + 2NaNO₃

With chemical reaction n(Pb(NO₃)₂) = ½ n(NaI), so 10 moles of sodium iodide (NaI) will interact with 5 moles of lead nitrate (Pb(NO₃)₂).

Answer: 5 moles.