What mass of precipitate will result from the following scenario?

Question

A $1.25 \cdot L$ $1.25L$ volume of $lead nitrate$ $"lead nitrate"$ at $0.050 \cdot m o l \cdot L^{- 1}$ $0.050molL^-1$ concentration is mixed with a $2.00 \cdot L$ $2.00L$ volume of $sodium sulfate$ $"sodium sulfate"$ at $0.025 \cdot m o l \cdot L^{- 1}$ $0.025molL^-1$ concentration. Write the stoichiometric equation, and determine the equivalence.

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Answer 1 · 2016-10-31T07:14:48

Approx, $15 \cdot g$ $15*g$ $P b S O_{4}$ $PbSO_4$ precipitate.

We need (i) a stoichiometric equation:

$P b^{2 +} + S O_{4}^{2 -} \to P b S O_{4} (s) ⏐ ⏐ ↓$ $Pb^(2+) + SO_4^(2-) rarr PbSO_4(s)darr$

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

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

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

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 · 2016-10-31T07:24:44

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

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