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20 June, 23:50

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

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Answers (2)
  1. 21 June, 01:36
    0
    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
  2. 21 June, 01:54
    0
    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)
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