How many moles of potassium iodide, KI, are required to precipitate all of the lead(II)ion from 50.0 mL of a 1.2 M Pb(NO₃)₂ solution? (First, write a balanced equation for the reaction.) A)0.12 mole B)0.060 mole C)0.24 mole D)0.030 mole E)0.048 mole

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Quizplus · Accepted Answer

The balanced equation for the reaction is:Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)From the balanced equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI. Therefore, the number of moles of KI required to precipitate all of the lead(II) ion from 50.0 mL of a 1.2 M Pb(NO3)2 solution is:(1.2 mol/L) x (50.0 mL/1000 mL/L) x (2 mol KI/1 mol Pb(NO3)2) = 0.12 mol KITherefore, the correct answer is A) 0.12 mole.

How Many Moles of Potassium Iodide, KI, Are Required to Precipitate