Question

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.

Answer 1

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?

Answer 2

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`