A-Level AQA Chemistry The Mole, Avogadro & The Ideal Gas Equation: Lead(II) nitrate and potassium i

Lead(II) nitrate and potassium iodide react according to the equation Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) In an experiment, 25.0 cm³ of a 0.100 mol dm⁻³ solution of each compound are mixed together - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 3

Question 7

Lead(II)-nitrate-and-potassium-iodide-react-according-to-the-equation--Pb(NO₃)₂(aq)-+-2KI(aq)-→-PbI₂(s)-+-2KNO₃(aq)--In-an-experiment,-25.0-cm³-of-a-0.100-mol-dm⁻³-solution-of-each-compound-are-mixed-together-AQA-A-Level Chemistry-Question 7-2019-Paper 3.png

Lead(II) nitrate and potassium iodide react according to the equation Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) In an experiment, 25.0 cm³ of a 0.100 mol dm⁻³ s... show full transcript

Worked Solution & Example Answer:Lead(II) nitrate and potassium iodide react according to the equation Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) In an experiment, 25.0 cm³ of a 0.100 mol dm⁻³ solution of each compound are mixed together - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 3

Step 1

Calculate the moles of Pb(NO₃)₂

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Answer

To find the moles of lead(II) nitrate, we use the formula:

ext{Moles} = ext{Concentration} imes ext{Volume}

Where:

Concentration = 0.100 mol dm⁻³
Volume = 25.0 cm³ = 0.0250 dm³ (since 1 dm³ = 1000 cm³)

Thus:

ext{Moles of Pb(NO₃)₂} = 0.100 imes 0.0250 = 0.00250 ext{ mol}

Step 2

Calculate the moles of KI

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Answer

Using the same formula for potassium iodide:

ext{Moles of KI} = 0.100 imes 0.0250 = 0.00250 ext{ mol}

Step 3

Determine the limiting reactant

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Answer

The balanced equation shows that 1 mole of Pb(NO₃)₂ reacts with 2 moles of KI. Therefore, we need:

For 0.00250 mol of Pb(NO₃)₂, we require:

0.00250 imes 2 = 0.00500 ext{ mol of KI}

Since we only have 0.00250 mol of KI, KI is the limiting reactant.

Step 4

Calculate the moles of PbI₂ produced

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Answer

From the balanced equation, 2 moles of KI produce 1 mole of PbI₂. Therefore:

ext{Moles of PbI₂} = rac{0.00250}{2} = 0.00125 ext{ mol}

Thus, the amount of lead(II) iodide formed is 1.25 x 10⁻³ mol.

Lead(II) nitrate and potassium iodide react according to the equation Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) In an experiment, 25.0 cm³ of a 0.100 mol dm⁻³ solution of each compound are mixed together - AQA - A-Level Chemistry - Question 7 - 2019 - Paper 3

Calculate the moles of Pb(NO₃)₂

Calculate the moles of KI

Determine the limiting reactant

Calculate the moles of PbI₂ produced

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