Question

What mass of precipitate can form if 1.5 L of a saturated solution of Pb(NO3)2 is mixed with 0.765 L of a 0.205 M NaI solution?

Answer 1

Answer:

$m_{PbI_2}=36.2gPbI_2$

Explanation:

Hello!

In this case, since the chemical reaction that takes place when Pb(NO3)2 and NaI are mixed is:

$Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)$

By which a solid precipitate of PbI2  is produced. In such a way, since we know the volume and molarity of NaI, we can compute the moles of NaI as shown below:

$n_{NaI}=0.765L*0.205\frac{molNaI}{L}=0.157molNaI$

Next, since it is in a 2:1 mole ratio with PbI2 (molar mass = 461.01  g/mol) we compute the mass of PbI2  precipitate as shown below:

$m_{PbI_2}=0.157molNaI*\frac{1molPbI_2}{2molNaI}*\frac{461.01 gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=36.2gPbI_2$

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