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When heated, a sample of potassium chlorate(V) (KClO₃) produced 67.2 cm³ of oxygen, measured at 298 K and 110 kPa - AQA - A-Level Chemistry - Question 7 - 2020 - Paper 3

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When heated, a sample of potassium chlorate(V) (KClO₃) produced 67.2 cm³ of oxygen, measured at 298 K and 110 kPa. 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) What is the amou... show full transcript

Worked Solution & Example Answer:When heated, a sample of potassium chlorate(V) (KClO₃) produced 67.2 cm³ of oxygen, measured at 298 K and 110 kPa - AQA - A-Level Chemistry - Question 7 - 2020 - Paper 3

Step 1

Calculate moles of oxygen produced

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Answer

To calculate the number of moles of oxygen produced, we can use the ideal gas law formula:

PV=nRTPV = nRT

Where:

  • P = pressure in pascals = 110 kPa = 110,000 Pa
  • V = volume in cubic meters = 67.2 cm³ = 67.2 × 10^{-6} m³
  • n = number of moles of gas
  • R = gas constant = 8.31 J K⁻¹ mol⁻¹
  • T = temperature in kelvin = 298 K

Substituting in the known values:

n=PVRT=(110000Pa)×(67.2×106m3)(8.31J K1mol1)×(298K)n = \frac{PV}{RT} = \frac{(110000 \, \text{Pa}) \times (67.2 \times 10^{-6} \, \text{m}^3)}{(8.31 \, \text{J K}^{-1} \text{mol}^{-1}) \times (298 \, \text{K})}

Calculating this gives:

n2.73×103 moles of O2n \approx 2.73 \times 10^{-3} \text{ moles of } O_2

Step 2

Determine moles of KClO₃ decomposed

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Answer

From the balanced equation, we observe that 3 moles of O₂ are produced from 2 moles of KClO₃:

2 KClO33 O22 \text{ KClO}_3 \rightarrow 3 \text{ O}_2

Therefore, we can set up a ratio to find the moles of KClO₃:

2 moles KClO33 moles O2\frac{2 \text{ moles KClO}_3}{3 \text{ moles O}_2}

Let (x) be the moles of KClO₃ decomposed:

x(2.73×103)=23\frac{x}{(2.73 \times 10^{-3})} = \frac{2}{3}

Solving for (x):

x=23×(2.73×103)1.82×103 moles of KClO3x = \frac{2}{3} \times (2.73 \times 10^{-3})\approx 1.82 \times 10^{-3} \text{ moles of KClO}_3

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