10 mL of 0.1 M pb(no3)2 is mixed with 20 mL of 0.02 M KI.the amount of pbI2 precipitate will be equal to

To determine the amount of lead(II) iodide (PbI2) precipitate formed when mixing 10 mL of 0.1 M Pb(NO3)2 with 20 mL of 0.02 M KI, we need to follow a few logical steps involving stoichiometry and the concept of limiting reagents.

Understanding the Reaction

The chemical reaction between lead(II) nitrate and potassium iodide can be represented as follows:

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

This equation indicates that one mole of lead(II) nitrate reacts with two moles of potassium iodide to produce one mole of lead(II) iodide precipitate and two moles of potassium nitrate.

Calculating Moles of Reactants

First, we need to calculate the number of moles of each reactant present in the solutions.

• For Pb(NO3)2:

Volume = 10 mL = 0.010 L

Concentration = 0.1 M

Moles of Pb(NO3)2 = Concentration × Volume = 0.1 mol/L × 0.010 L = 0.001 moles

• For KI:

Volume = 20 mL = 0.020 L

Concentration = 0.02 M

Moles of KI = Concentration × Volume = 0.02 mol/L × 0.020 L = 0.0004 moles

Identifying the Limiting Reagent

Next, we need to determine which reactant is the limiting reagent. According to the balanced equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI. Therefore, we can calculate how much KI is needed to react with the available Pb(NO3)2:

For 0.001 moles of Pb(NO3)2, the required moles of KI = 0.001 moles Pb(NO3)2 × 2 moles KI/mole Pb(NO3)2 = 0.002 moles KI.

Since we only have 0.0004 moles of KI available, KI is the limiting reagent.

Calculating the Amount of PbI2 Formed

Now that we know KI is the limiting reagent, we can find out how much PbI2 will be produced. According to the balanced equation, 2 moles of KI produce 1 mole of PbI2. Therefore, the moles of PbI2 produced from 0.0004 moles of KI is:

Moles of PbI2 = 0.0004 moles KI × (1 mole PbI2 / 2 moles KI) = 0.0002 moles PbI2.

Calculating the Mass of PbI2 Precipitate

To find the mass of the PbI2 precipitate, we need to use its molar mass. The molar mass of PbI2 can be calculated as follows:

• Lead (Pb) = 207.2 g/mol
• Iodine (I) = 126.9 g/mol × 2 = 253.8 g/mol
• Total Molar Mass of PbI2 = 207.2 g/mol + 253.8 g/mol = 461 g/mol

Now, we can calculate the mass of PbI2 formed:

Mass of PbI2 = Moles × Molar Mass = 0.0002 moles × 461 g/mol = 0.0922 g.

Final Result

Therefore, the amount of PbI2 precipitate formed when 10 mL of 0.1 M Pb(NO3)2 is mixed with 20 mL of 0.02 M KI is approximately 0.0922 grams.