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How many liters of 0.150 m ki would be required to react

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Reference no: EM13498479

How many liters of 0.150 M kI would be required to react completely with 15.00 grams of lead (II) nitrate?

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

Reference no: EM13498479

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