Question

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?

Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answer 1

`m_(PbI_2)=18.2gKI`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2`

Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:

`n_(Pb(NO_3)_2)=96.7mL*(1.134g)/(mL)*(0.14gPb(NO_3)_2)/(1g)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) =0.0464molPb(NO_3)_2\\\\n_(KI)=99.8mL*(1.093g)/(mL)*(0.12gKI)/(1g)*(1molKI)/(166gKI) =0.0789molKI`

Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):

`n_(KI)^(consumed\ by\ Pb(NO_3)_2)=0.0464molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0928molKI`

Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:

`m_(PbI_2)=0.0789molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \\\\m_(PbI_2)=18.2gKI`

Best regards.