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?

Question

Cameron Andrews · Accepted Answer

HCl has a volume of 90.7 mL.

The equation that is balanced is:

Pb(NO₃)₂ + 2HCl → PbCl₂+2HNO₃

You have to change.

Mole of PbCl₂ → moles of HCl → litres of HCl from grams of PbCl₂

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

Thus, to precipitate 1.64 g of PbCl₂, 90.7 mL of 0.130 mol/L HCl is required.

Caleb Adams · Answer

undefined First, write the balanced chemical equation:

Pb(NO3)2(aq) + 2HCl(aq) → PbCl2(s) + 2HNO3(aq)

The molar mass of PbCl2 is 278.1 g/mol.

Calculate moles of PbCl2:

$$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $$

Calculate moles of HCl:

$$ \text{moles} = \frac{\text{moles of PbCl2}}{2} $$

Calculate volume of HCl:

$$ \text{volume} = \frac{\text{moles of HCl}}{\text{concentration}} $$