Question

A 1.30g sample of Lead(II) Hydroxide was reacted with 50 mL of 0.250 mol/L Hydrochloric Acid to form a solid precipitate. How much solid precipitate should be formed? (Ans: 1.50g)

Answer 1

AS with all these sorts of problems we first write a stoichiometric equation.....

Explanation:

`Pb(OH)_2(s) +2HCl(aq) rarr PbCl_2(s) + 2H_2O(l)`

...we write the equation so as to establish the stoichiometric equivalence.

`"Moles of lead salt"=(1.30*g)/(241.21*g*mol^-1)=5.39xx10^-3*mol`

`"Moles of acid"=50*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=0.0125*mol`.

Chloride ion is thus present in stoichiometric excess....and the (insoluble) lead salt should quantitatively precipitate as a white solid....

`"Mass of lead chloride"=5.39xx10^-3*molxx278.10*g*mol^-1=1.50*g`....as required....