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For each precipitation reaction, calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant.

2KI(aq) + Pb(NO3)2(aq)--->PbI2(s) + 2KNO3(aq)

Not looking for the answer I really want to know if I did it correctly

17.3g Pb(NO3)/331.22gPb(NO3) x 2 mol KI x 166g KI = 17.3G KI

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It looks ok to me including the answer.

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