Question
What volume of oxygen at stp will be obtained when 24.5g of potassium trioxochlorate (5) crystals are heated to give potassium chlorate and oxygen
Answers
GPT-4o mini
To find the volume of oxygen produced when heating potassium trioxochlorate (V) \((KClO_3)\), we first need to understand the reaction that occurs when it is heated. The decomposition of potassium trioxochlorate (V) can be represented by the following balanced chemical equation:
\[
2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g)
\]
From this equation, we can see that 2 moles of potassium trioxochlorate decompose to produce 3 moles of oxygen gas.
Next, we need to:
1. Calculate the molar mass of potassium trioxochlorate (V) \((KClO_3)\).
- K: 39.1 g/mol
- Cl: 35.5 g/mol
- O: 16.0 g/mol (oxygen, 3 atoms)
Molar mass of \(KClO_3 = 39.1 + 35.5 + (3 \times 16.0) = 39.1 + 35.5 + 48.0 = 122.6 \, g/mol\)
2. Determine the number of moles of \(KClO_3\) in 24.5 g:
\[
\text{Moles of } KClO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{24.5 \, g}{122.6 \, g/mol} \approx 0.200 \, mol
\]
3. Use the stoichiometry from the balanced equation to find the moles of oxygen produced. According to the reaction, 2 moles of \(KClO_3\) produce 3 moles of \(O_2\). Therefore, 0.200 moles of \(KClO_3\) will produce:
\[
\text{Moles of } O_2 = 0.200 \, mol \times \frac{3 \, mol \, O_2}{2 \, mol \, KClO_3} = 0.300 \, mol \, O_2
\]
4. Now we can calculate the volume of oxygen gas produced at STP (Standard Temperature and Pressure). At STP, 1 mole of gas occupies 22.4 liters:
\[
\text{Volume of } O_2 = 0.300 \, mol \times 22.4 \, L/mol = 6.72 \, L
\]
Thus, the volume of oxygen produced when 24.5 g of potassium trioxochlorate (V) is heated to decomposition is approximately **6.72 liters**.
\[
2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g)
\]
From this equation, we can see that 2 moles of potassium trioxochlorate decompose to produce 3 moles of oxygen gas.
Next, we need to:
1. Calculate the molar mass of potassium trioxochlorate (V) \((KClO_3)\).
- K: 39.1 g/mol
- Cl: 35.5 g/mol
- O: 16.0 g/mol (oxygen, 3 atoms)
Molar mass of \(KClO_3 = 39.1 + 35.5 + (3 \times 16.0) = 39.1 + 35.5 + 48.0 = 122.6 \, g/mol\)
2. Determine the number of moles of \(KClO_3\) in 24.5 g:
\[
\text{Moles of } KClO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{24.5 \, g}{122.6 \, g/mol} \approx 0.200 \, mol
\]
3. Use the stoichiometry from the balanced equation to find the moles of oxygen produced. According to the reaction, 2 moles of \(KClO_3\) produce 3 moles of \(O_2\). Therefore, 0.200 moles of \(KClO_3\) will produce:
\[
\text{Moles of } O_2 = 0.200 \, mol \times \frac{3 \, mol \, O_2}{2 \, mol \, KClO_3} = 0.300 \, mol \, O_2
\]
4. Now we can calculate the volume of oxygen gas produced at STP (Standard Temperature and Pressure). At STP, 1 mole of gas occupies 22.4 liters:
\[
\text{Volume of } O_2 = 0.300 \, mol \times 22.4 \, L/mol = 6.72 \, L
\]
Thus, the volume of oxygen produced when 24.5 g of potassium trioxochlorate (V) is heated to decomposition is approximately **6.72 liters**.