Question

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)=18.2gPbI_2`

Step-by-step explanation:

Hello,

In this case, we write the reaction again:

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

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

`n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\\\\n_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\\\\n_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI) *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\\\\n_(KI)=0.07885molKI`

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

`0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI`

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

`m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \\\\m_(PbI_2)=18.2gPbI_2`

Best regards.

Answer 2

Mass PbI2 = 18.19 grams

Step-by-step explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

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

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams