# What mass of precipitate will result from the following scenario?

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

Oct 31, 2016

Approx, $15 \cdot g$ $P b S {O}_{4}$ precipitate.

#### Explanation:

We need (i) a stoichiometric equation:

$P {b}^{2 +} + S {O}_{4}^{2 -} \rightarrow P b S {O}_{4} \left(s\right) \downarrow$

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

$\text{Moles of lead}$ $=$ $1.25 \cdot \cancel{L} \times 0.0500 \cdot m o l \cdot \cancel{{L}^{-} 1} = 0.0625 \cdot m o l$.

$\text{Moles of sulfate}$ $=$ $2.0 \cdot \cancel{L} \times 0.0250 \cdot m o l \cdot \cancel{{L}^{-} 1} = 0.0500 \cdot m o l$.

Clearly, sulfate is the limiting reagent. And according to the given stoichiometry, $0.0500 \cdot m o l$ of $P b S {O}_{4}$ will crash out of solution.

$\text{Mass of lead sulfate}$ $=$ 0.0500*molxx303.26*g*mol^-1=??

What does $\text{stoichiometric}$ mean?

Oct 31, 2016

1.25L of 0.0500 M $P b {\left(N {O}_{3}\right)}_{2}$

$\equiv 1.25 L \times 0.05 \frac{\text{mol}}{L} = 6.25 \times {10}^{-} 2 m o l$

and

Similarly 2.00L of 0.0250 M $N {a}_{2} S {O}_{4}$

$\equiv 2 \times 0.025 m o l = 5 \times {10}^{-} 2 m o l$

Balanced equation

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + N {a}_{2} S {O}_{4} \left(a q\right) \to P b S {O}_{4} \left(s\right) \downarrow + 2 N a N {O}_{3} \left(a q\right)$

This equation suggests that two reactants react 1 mole each to produce 1mole $P b S {O}_{4}$

Here limiting reagent is $N {a}_{2} S {O}_{4}$ and $5.0 \times {10}^{-} 2$ moles of it reacts with same no. of mole of $P b {\left(N {O}_{3}\right)}_{2}$ to produce $5 \times {10}^{-} 2 m o l \text{ } P b S {O}_{4}$

Molar mass of $P b S {O}_{4}$

$= 207 + 32 + 4 \cdot 16 = 303 \text{ g/mol}$

So mass of $P b S {O}_{4}$ precipitated wil be

$= 5 \times {10}^{\text{-2"molxx303g/"mol}} = 15.15 g$