chemistry homework

Question

How many moles of potassium chloride are needed to precipitate all the lead(II) ion from 14.0 mL of a 0.350 M lead(II) nitrate solution?

Answer 1

2KCl(aq) + Pb²⁺(aq) ==> PbCl₂(s) + 2K⁺(aq) ... balanced equation

mols Pb²⁺ = 14.0 ml Pb(NO₃)₂ x 1L/1000 ml x 0.350 mol/L x 1 mol Pb²⁺/mol Pb(NO₃)₂ = 0.00490 mols Pb²⁺

mols KCl needed = 0.00490 mol Pb²⁺ + 2 mol KCl/mol Pb²⁺ = 0.00980 mols KCl needed