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