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If 229 g of Pb(NO3)2 and 190 mL of 1.68 M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many grams of Pb(NO3)2 remain? Round your answer to 3 significant figures.
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)

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    Sammy, Todd, John, whomever. This problem is almost the same as the last one. You try it on your own. Post your work if you stuck.

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