What mass of precipitate will result from the following scenario?

A #1.25*L# volume of #"lead nitrate"# at #0.050*mol*L^-1# concentration is mixed with a #2.00*L# volume of #"sodium sulfate"# at #0.025*mol*L^-1# concentration. Write the stoichiometric equation, and determine the equivalence.

2 Answers
Oct 31, 2016

Answer:

Approx, #15*g# #PbSO_4# precipitate.

Explanation:

We need (i) a stoichiometric equation:

#Pb^(2+) + SO_4^(2-) rarr PbSO_4(s)darr#

And (ii) the stoichiometric quantities of the reagents involved.

#"Moles of lead"# #=# #1.25*cancelLxx0.0500*mol*cancel(L^-1)=0.0625*mol#.

#"Moles of sulfate"# #=# #2.0*cancelLxx0.0250*mol*cancel(L^-1)=0.0500*mol#.

Clearly, sulfate is the limiting reagent. And according to the given stoichiometry, #0.0500*mol# of #PbSO_4# will crash out of solution.

#"Mass of lead sulfate"# #=# #0.0500*molxx303.26*g*mol^-1=??#

What does #"stoichiometric"# mean?

Oct 31, 2016

1.25L of 0.0500 M #Pb(NO_3)_2#

#equiv1.25Lxx0.05"mol"/L=6.25xx10^-2mol#

and

Similarly 2.00L of 0.0250 M #Na_2SO_4#

#equiv2xx0.025mol=5xx10^-2mol#

Balanced equation

#Pb(NO_3)_2(aq) + Na_2SO_4(aq) ->PbSO_4(s)darr+2NaNO_3(aq)#

This equation suggests that two reactants react 1 mole each to produce 1mole #PbSO_4#

Here limiting reagent is #Na_2SO_4# and #5.0xx10^-2# moles of it reacts with same no. of mole of #Pb(NO_3)_2# to produce #5xx10^-2mol" "PbSO_4#

Molar mass of #PbSO_4#

#=207+32+4*16=303" g/mol"#

So mass of #PbSO_4# precipitated wil be

#=5xx10^"-2"molxx303g/"mol"=15.15g#