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If 250 mL of 0.375 mol/L solution of potassium iodide was mixed with an excess of lead (II) nitrate, what mass of lead (II) iodide precipitate would form? Please tell me the steps to solve problem- not
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If 250 mL of 0.375 mol/L solution of potassium iodide was mixed with an excess of lead (II) nitrate, what mass of lead (II) iodide precipitate would form? Please tell me the steps to solve problem- not just the answer :)

User IUnknown
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1 Answer

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The balanced chemical equation is Pb(NO3)2+ 2KI produces PbI2 + 2K(NO)3

.4 L of KI × (.375 mol/L of KI) × (1 mol of PbI2 / 2 mol of KI) × (461 g of PbI2/1 mol of PbI2)

=34. 6 g of PbI2 precipitate.
User Rudigrobler
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