A-Level AQA Chemistry Formulae, Equations & Calculations: 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 moles of lead(II) nitrate

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Answer

First, we need to calculate the moles of lead(II) nitrate (Pb(NO₃)₂) present in the solution.

Given:

Volume of Pb(NO₃)₂ solution = 25.0 cm³ = 0.025 dm³
Concentration = 0.100 mol dm⁻³

Using the formula for moles:

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

We have:

ext{Moles of Pb(NO₃)₂} = 0.100 ext{ mol dm}^{-3} imes 0.025 ext{ dm}^{3} = 0.0025 ext{ mol}

Step 2

Calculate moles of potassium iodide

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Answer

Next, we calculate the moles of potassium iodide (KI) in the solution. The same volume and concentration apply:

ext{Moles of KI} = 0.100 ext{ mol dm}^{-3} imes 0.025 ext{ dm}^{3} = 0.0025 ext{ mol}

Step 3

Determine the limiting reagent

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Answer

According to the balanced chemical equation, 1 mole of Pb(NO₃)₂ reacts with 2 moles of KI. Therefore:

Moles of KI required for 0.0025 moles of Pb(NO₃)₂ = 2 × 0.0025 = 0.0050 moles

Since we only have 0.0025 moles of KI, it is the limiting reagent.

Step 4

Calculate moles of lead(II) iodide formed

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Answer

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

Moles of PbI₂ formed = 0.0025 moles of KI × (1 mol PbI₂ / 2 mol KI) = 0.00125 moles.

Converting this to scientific notation gives:

ext{Moles of PbI₂} = 1.25 imes 10^{-3} ext{ 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 moles of lead(II) nitrate

Calculate moles of potassium iodide

Determine the limiting reagent

Calculate moles of lead(II) iodide formed

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