Answer :
To determine the mass of oxygen gas formed from the decomposition of potassium chlorate (KClO₃), we can follow a step-by-step process.
1. Write the Balanced Chemical Equation:
[tex]\( 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \)[/tex]
2. Identify the Given Data:
- Mass of KClO₃: 240 grams
- Molar mass of KClO₃: 122.5 g/mol
- Molar mass of O₂: 31.998 g/mol
3. Calculate the Moles of KClO₃:
The number of moles of KClO₃ can be calculated using its mass and molar mass:
[tex]\[ \text{moles of KClO₃} = \frac{\text{mass of KClO₃}}{\text{molar mass of KClO₃}} = \frac{240 \, \text{g}}{122.5 \, \text{g/mol}} \][/tex]
After calculation:
[tex]\[ \text{moles of KClO₃} \approx 1.959 \, \text{mol} \][/tex]
4. Use Stoichiometry to Find Moles of O₂ Formed:
According to the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂:
[tex]\[ \text{moles of O₂} = \left(\frac{3}{2}\right) \times \text{moles of KClO₃} = \left(\frac{3}{2}\right) \times 1.959 \, \text{mol} \][/tex]
After calculation:
[tex]\[ \text{moles of O₂} \approx 2.9385 \, \text{mol} \][/tex]
5. Convert Moles of O₂ to Mass of O₂:
The mass of O₂ can be found using its molar mass:
[tex]\[ \text{mass of O₂} = \text{moles of O₂} \times \text{molar mass of O₂} = 2.9385 \, \text{mol} \times 31.998 \, \text{g/mol} \][/tex]
After calculation:
[tex]\[ \text{mass of O₂} \approx 94.080 \, \text{g} \][/tex]
Therefore, the correct calculation that corresponds to determining the mass of oxygen gas formed is:
[tex]\[ (240 \, \text{g}) \times \left(\frac{3 \times 31.998 \, \text{g/mol}}{2 \times 122.5 \, \text{g/mol}}\right) \][/tex]
Reviewing the options given:
- [tex]\((240 \times 2 \times 31.998)+(122.5 \times 3) \, \text{g}\)[/tex] does not correctly follow the stoichiometry or the unit conversions required.
- [tex]\((240 \times 3 \times 31.998)+(122.5 \times 2) \, \text{g}\)[/tex] also does not match the stoichiometry process.
- [tex]\((240 \times 2 \times 122.5) \div(31.998 \times 3) \, \text{g}\)[/tex] seems to have a mix-up in the placement of molar masses, leading to incorrect units.
- [tex]\((240 \times 3 \times 122.5)+(31.998 \times 2) \, \text{g}\)[/tex] does not follow the correct process at all.
Thus, the correct step-by-step approach which we followed gives the mass of oxygen gas correctly, and none of the options provided match exactly with the logical calculation process we followed. But following correct steps and the method ensures the correct mass calculation for O₂.
For understanding purpose, the closest that follows a somewhat logical process would involve a mixture of the molar masses feasibly converting units properly as detailed in the manual solution.
1. Write the Balanced Chemical Equation:
[tex]\( 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \)[/tex]
2. Identify the Given Data:
- Mass of KClO₃: 240 grams
- Molar mass of KClO₃: 122.5 g/mol
- Molar mass of O₂: 31.998 g/mol
3. Calculate the Moles of KClO₃:
The number of moles of KClO₃ can be calculated using its mass and molar mass:
[tex]\[ \text{moles of KClO₃} = \frac{\text{mass of KClO₃}}{\text{molar mass of KClO₃}} = \frac{240 \, \text{g}}{122.5 \, \text{g/mol}} \][/tex]
After calculation:
[tex]\[ \text{moles of KClO₃} \approx 1.959 \, \text{mol} \][/tex]
4. Use Stoichiometry to Find Moles of O₂ Formed:
According to the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂:
[tex]\[ \text{moles of O₂} = \left(\frac{3}{2}\right) \times \text{moles of KClO₃} = \left(\frac{3}{2}\right) \times 1.959 \, \text{mol} \][/tex]
After calculation:
[tex]\[ \text{moles of O₂} \approx 2.9385 \, \text{mol} \][/tex]
5. Convert Moles of O₂ to Mass of O₂:
The mass of O₂ can be found using its molar mass:
[tex]\[ \text{mass of O₂} = \text{moles of O₂} \times \text{molar mass of O₂} = 2.9385 \, \text{mol} \times 31.998 \, \text{g/mol} \][/tex]
After calculation:
[tex]\[ \text{mass of O₂} \approx 94.080 \, \text{g} \][/tex]
Therefore, the correct calculation that corresponds to determining the mass of oxygen gas formed is:
[tex]\[ (240 \, \text{g}) \times \left(\frac{3 \times 31.998 \, \text{g/mol}}{2 \times 122.5 \, \text{g/mol}}\right) \][/tex]
Reviewing the options given:
- [tex]\((240 \times 2 \times 31.998)+(122.5 \times 3) \, \text{g}\)[/tex] does not correctly follow the stoichiometry or the unit conversions required.
- [tex]\((240 \times 3 \times 31.998)+(122.5 \times 2) \, \text{g}\)[/tex] also does not match the stoichiometry process.
- [tex]\((240 \times 2 \times 122.5) \div(31.998 \times 3) \, \text{g}\)[/tex] seems to have a mix-up in the placement of molar masses, leading to incorrect units.
- [tex]\((240 \times 3 \times 122.5)+(31.998 \times 2) \, \text{g}\)[/tex] does not follow the correct process at all.
Thus, the correct step-by-step approach which we followed gives the mass of oxygen gas correctly, and none of the options provided match exactly with the logical calculation process we followed. But following correct steps and the method ensures the correct mass calculation for O₂.
For understanding purpose, the closest that follows a somewhat logical process would involve a mixture of the molar masses feasibly converting units properly as detailed in the manual solution.
