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

`m_(PbI_2)=36.2gPbI_2`

Step-by-step 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(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*(1molPbI_2)/(2molNaI)*(461.01 gPbI_2)/(1molPbI_2) \\\\m_(PbI_2)=36.2gPbI_2`

Best regards!