Answer :
Sure, let's break down the solution step-by-step to determine how many grams of lead (II) oxide (PbO) will be produced by the decomposition of 2.50 grams of lead (II) carbonate (PbCO₃).
1. Write down the given data:
- Mass of PbCO₃: 2.50 grams
2. Determine the molar masses of involved compounds:
- Molar mass of PbCO₃ [tex]\( = \)[/tex] Molar mass of Pb [tex]\( + \)[/tex] Molar mass of C [tex]\( + \)[/tex] 3 [tex]\( \times \)[/tex] Molar mass of O
- Molar mass of PbCO₃ [tex]\( = \)[/tex] 207.2 g/mol (Pb) [tex]\( + \)[/tex] 12.0 g/mol (C) [tex]\( + \)[/tex] 3 [tex]\( \times \)[/tex] 16.0 g/mol (O)
- Molar mass of PbCO₃ [tex]\( = \)[/tex] 207.2 + 12.0 + 48.0 [tex]\( = \)[/tex] 267.2 g/mol
3. Calculate the number of moles of PbCO₃ decomposed:
- Moles of PbCO₃ [tex]\( = \)[/tex] [tex]\(\frac{\text{Mass of PbCO₃}}{\text{Molar mass of PbCO₃}}\)[/tex]
- Moles of PbCO₃ [tex]\( = \)[/tex] [tex]\(\frac{2.50 \text{ grams}}{267.2 \text{ g/mol}}\)[/tex]
- Moles of PbCO₃ [tex]\( \approx \)[/tex] 0.00936 moles
4. From the balanced equation [tex]\( PbCO₃ \rightarrow PbO + CO₂ \)[/tex], we see that:
- 1 mole of PbCO₃ produces 1 mole of PbO.
- Therefore, the moles of PbO produced will be equal to the moles of PbCO₃ decomposed.
5. Calculate the molar mass of PbO:
- Molar mass of PbO [tex]\( = \)[/tex] Molar mass of Pb [tex]\( + \)[/tex] Molar mass of O
- Molar mass of PbO [tex]\( = \)[/tex] 207.2 g/mol (Pb) [tex]\( + \)[/tex] 16.0 g/mol (O)
- Molar mass of PbO [tex]\( = \)[/tex] 223.2 g/mol
6. Calculate the mass of PbO produced:
- Mass of PbO [tex]\( = \)[/tex] Moles of PbO [tex]\( \times \)[/tex] Molar mass of PbO
- Mass of PbO [tex]\( = \)[/tex] 0.00936 moles [tex]\( \times \)[/tex] 223.2 g/mol
- Mass of PbO [tex]\( \approx \)[/tex] 2.09 grams
7. Conclusion:
- The mass of lead (II) oxide produced from the decomposition of 2.50 grams of lead (II) carbonate is approximately 2.09 grams.
So, the correct answer is:
e. 2.09
1. Write down the given data:
- Mass of PbCO₃: 2.50 grams
2. Determine the molar masses of involved compounds:
- Molar mass of PbCO₃ [tex]\( = \)[/tex] Molar mass of Pb [tex]\( + \)[/tex] Molar mass of C [tex]\( + \)[/tex] 3 [tex]\( \times \)[/tex] Molar mass of O
- Molar mass of PbCO₃ [tex]\( = \)[/tex] 207.2 g/mol (Pb) [tex]\( + \)[/tex] 12.0 g/mol (C) [tex]\( + \)[/tex] 3 [tex]\( \times \)[/tex] 16.0 g/mol (O)
- Molar mass of PbCO₃ [tex]\( = \)[/tex] 207.2 + 12.0 + 48.0 [tex]\( = \)[/tex] 267.2 g/mol
3. Calculate the number of moles of PbCO₃ decomposed:
- Moles of PbCO₃ [tex]\( = \)[/tex] [tex]\(\frac{\text{Mass of PbCO₃}}{\text{Molar mass of PbCO₃}}\)[/tex]
- Moles of PbCO₃ [tex]\( = \)[/tex] [tex]\(\frac{2.50 \text{ grams}}{267.2 \text{ g/mol}}\)[/tex]
- Moles of PbCO₃ [tex]\( \approx \)[/tex] 0.00936 moles
4. From the balanced equation [tex]\( PbCO₃ \rightarrow PbO + CO₂ \)[/tex], we see that:
- 1 mole of PbCO₃ produces 1 mole of PbO.
- Therefore, the moles of PbO produced will be equal to the moles of PbCO₃ decomposed.
5. Calculate the molar mass of PbO:
- Molar mass of PbO [tex]\( = \)[/tex] Molar mass of Pb [tex]\( + \)[/tex] Molar mass of O
- Molar mass of PbO [tex]\( = \)[/tex] 207.2 g/mol (Pb) [tex]\( + \)[/tex] 16.0 g/mol (O)
- Molar mass of PbO [tex]\( = \)[/tex] 223.2 g/mol
6. Calculate the mass of PbO produced:
- Mass of PbO [tex]\( = \)[/tex] Moles of PbO [tex]\( \times \)[/tex] Molar mass of PbO
- Mass of PbO [tex]\( = \)[/tex] 0.00936 moles [tex]\( \times \)[/tex] 223.2 g/mol
- Mass of PbO [tex]\( \approx \)[/tex] 2.09 grams
7. Conclusion:
- The mass of lead (II) oxide produced from the decomposition of 2.50 grams of lead (II) carbonate is approximately 2.09 grams.
So, the correct answer is:
e. 2.09
