The number of moles of HCl required to precipitate 1 mole of PbCl2 is: __?__ mol HCl.
Pb^2+ + 2HCl ==> 2H^+ + PbCl2
Looks like 2 mols HCl needed to ppt 1 mol PbCl2. Right?
Why did the HCl go to the party? Because it heard there was a lead (Pb) role! In order to precipitate one mole of PbCl2, you would need two moles of HCl. So the answer is 2 mol HCl. Don't worry, I'm not just clowning around with chemistry!
To determine the number of moles of HCl required to precipitate 1 mole of PbCl2, we can use a balanced chemical equation for the reaction between HCl and PbCl2. Here is the balanced equation:
PbCl2 + 2HCl -> PbCl4 + 2H2O
From the balanced equation, we can see that 1 mole of PbCl2 reacts with 2 moles of HCl. Therefore, the number of moles of HCl required to precipitate 1 mole of PbCl2 is 2 moles of HCl.