Calculate the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2

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To calculate the minimum number of moles of Pb(NO3)2 required to start precipitation, we need to determine the stoichiometry of the reaction between Pb(NO3)2 and ZnI2.

The balanced chemical equation for the reaction between Pb(NO3)2 and ZnI2 is:

Pb(NO3)2 + 2ZnI2 -> PbI2 + 2Zn(NO3)2

From the balanced chemical equation, we can see that one mole of Pb(NO3)2 reacts with 2 moles of ZnI2 to form one mole of PbI2.

Given:
- Volume of ZnI2 solution: 50.0 mL
- Concentration of ZnI2 solution: 0.15 M

To calculate the minimum number of moles of Pb(NO3)2 required, we need to use the stoichiometry of the reaction.

Step 1: Convert the volume of ZnI2 solution to liters:
50.0 mL = 50.0 / 1000 = 0.050 L

Step 2: Calculate the moles of ZnI2 in the solution:
Moles of ZnI2 = Volume (L) x Concentration (M)
= 0.050 L x 0.15 M
= 0.0075 moles

Step 3: Use the stoichiometry of the reaction to determine the moles of Pb(NO3)2 required:
From the balanced equation, we know that 2 moles of ZnI2 react with 1 mole of Pb(NO3)2.
Therefore, the moles of Pb(NO3)2 required = 0.0075 moles / 2 = 0.00375 moles.

Therefore, the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0 mL of 0.15M ZnI2 is 0.00375 moles.