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You mix 265.0 mL of 1.20 M lead(II) nitrate with 293 mL of 1.55 M potassium iodide. The lead(II) iodide is insoluble. What amount (g) of solid is produced from this mixture?

Answer :

Answer:

104.6 grams of PbI2 (solid) will be produced

Explanation:

Step 1: Data given

Volume of a 1.20 M lead(II) nitrate = 265.0 mL = 0.265 L

Volume of a 1.55 M potassium iodide = 293.0 mL = 0.293 L

Step 2: The balanced equation

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

Step 3: Calculate moles

Moles = molarity * volume

Moles Pb(NO3)2 = 1.20 M * 0.265 L

Moles Pb(NO3)2 = 0.318 moles

Moles KI = 1.55 M * 0.293 L

Moles KI = 0.454 moles

Step 4: Calculate the limiting reactant

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

Ki is the limiting reactant. It will completely be consumed. (0.454 moles) Pb(NO3)2 is in excess. There will react 0.454 / 2 = 0.227 moles

There will remain 0.318 moles - 0.227 =0.091 moles

Step 5: Calculete moles PbI2

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

For 0.454 moles KI we'll have 0.454/2 = 0.227 moles PbI2

Step 6: Calculate mass PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.227 moles * 461.01 g/mol

Mass PbI2 = 104.6 grams

104.6 grams of PbI2 (solid) will be produced

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Rewritten by : Barada

Answer:

105 grams PbI₂

Explanation:

Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)

moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole

moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant

moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂

grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)