How many moles of potassium iodide, KI, are required to precipitate all of the lead(II) ion from 25.0 mL of a 1.6 M Pb(NO3)2 solution? A. 0.020 mol B. 0.040 mol C. 0.080 mol D. 0.64 mol E. 64 mol
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To determine the number of moles of potassium iodide (KI) required to precipitate all of the lead(II) ions (Pb2+) from the solution, we need to use the balanced chemical equation for the reaction between Pb(NO3)2 and KI.
The balanced equation is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
From the equation, we can see that for every 1 mole of Pb(NO3)2, we need 2 moles of KI to fully react.
First, we need to calculate the number of moles of Pb(NO3)2 present in the solution. We can use the given concentration (1.6 M) and volume (25.0 mL) to do this.
Moles of Pb(NO3)2 = (concentration) x (volume)
= 1.6 M x 0.0250 L (converting mL to L)
= 0.04 moles
Since 1 mole of Pb(NO3)2 requires 2 moles of KI for complete reaction, we can now determine the number of moles of KI needed:
Moles of KI = (moles of Pb(NO3)2) x (2 moles KI / 1 mole Pb(NO3)2)
= 0.04 moles x (2 moles KI / 1 mole Pb(NO3)2)
= 0.08 moles
Therefore, the correct answer is C. 0.080 mol.