Question

What volume of 0.130 mol/L hydrochloric acid do you need to precipitate 1.64 g of lead(II) chloride from a solution of lead(II) nitrate?

Answer 1

The volume of HCl is 90.7 mL.

The balanced equation is:

2HCl + Pb(NO₃)₂ → PbCl₂+2HNO₃

You must convert

grams of PbCl₂ → moles of PbCl₂ → moles of HCl → litres of HCl

Volume of HCl = 1.64 g PbCl₂ × `(1"mol PbCl₂")/(278.1"g PbCl₂") × (2"mol HCl")/(1"mol PbCl₂") × (1"L HCl")/(0.130"mol HCl")` = 0.0907 L HCl = 90.7 mL HCl

So you need 90.7 mL of 0.130 mol/L HCl to precipitate 1.64 g of PbCl₂.