If 0.490 mol of Pb(NO3)2 and 130 mL of 1.08 M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many moles of Pb(NO3)2 remain? Round your answer to 3 significant figures.

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)

See below.