Question

When potassium chlorate is subjected to high temperatures, it decomposes into potassium chloride and oxygen. (a) Write a balanced equation for the decomposition. (b) In this decomposition, the actual yield is \(83.2 \%\). If \(198.5 \mathrm{~g}\) of oxygen are produced, how much potassium chlorate decomposed?

Short answer

Question: Write a balanced equation for the decomposition of potassium chlorate into potassium chloride and oxygen, and calculate how much potassium chlorate decomposed if the actual yield is \(83.2\%\) and \(198.5\mathrm{~g}\) of oxygen are produced. Answer: (a) The balanced equation for the decomposition is: 2 KClO\(_3\) → 2 KCl + 3 O\(_2\) (b) \(609\mathrm{~g}\) of potassium chlorate decomposed.

Step-by-step solution

Write unbalanced chemical equation

For the decomposition of potassium chlorate into potassium chloride and oxygen, we can write the unbalanced chemical equation as: KClO\(_3\) → KCl + O\(_2\)

Balance the chemical equation

To balance the chemical equation, we need to have the same number of atoms of each element on both sides of the equation. For this reaction, we can balance the equation as follows: 2 KClO\(_3\) → 2 KCl + 3 O\(_2\)

Calculate the moles of oxygen produced

First, we need to convert the mass of oxygen produced into moles. We can use the molar mass of oxygen (O\(_2\)) which is \(32 \mathrm{~g/mol}\). Moles of O\(_2\) = \(\frac{198.5\mathrm{~g}}{32 \mathrm{~g/mol}} = 6.203125\mathrm{~mol}\)

Determine moles of potassium chlorate decomposed

Using the balanced equation, we can determine the moles of potassium chlorate (KClO\(_3\)) decomposed by using stoichiometry. Since 2 moles of KClO\(_3\) produce 3 moles of O\(_2\), we can find the amount of KClO\(_3\) decomposed as follows: Moles of KClO\(_3\) = \(\frac{2}{3} \times 6.203125\mathrm{~mol} = \frac{2}{3} \times 6.203125 = 4.135417\mathrm{~mol}\)

Calculate the amount of potassium chlorate decomposed

We are given that the actual yield is \(83.2 \%\), so now we can find the actual moles of potassium chlorate decomposed using the percentage. Moles of KClO\(_3\) decomposed = \(\frac{4.135417\mathrm{~mol}}{0.832} = 4.972222\mathrm{~mol}\) Finally, we need to convert moles of potassium chlorate decomposed into grams by multiplying with its molar mass (which is \(39.10 + 35.45 + 3\times 16.00 = 122.55 \mathrm{~g/mol}\) for KClO\(_{3}\)). Amount of KClO\(_3\) decomposed = \(4.972222\mathrm{~mol} \times 122.55 \mathrm{~g/mol} = 609\mathrm{~g}\) In conclusion, (a) The balanced equation for the decomposition is: 2 KClO\(_3\) → 2 KCl + 3 O\(_2\) (b) \(609\mathrm{~g}\) of potassium chlorate decomposed.