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### Given Data:
1. Reaction:
[tex]\(2 \text{NaHCO}_3 \to \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\)[/tex]
2. Molar masses:
- [tex]\(M_r \text{of NaHCO}_3 = 84 \, \text{g/mol}\)[/tex]
- [tex]\(M_r \text{of Na}_2\text{CO}_3 = 106 \, \text{g/mol}\)[/tex]
3. Experimental data:
- Initial mass of sodium hydrogencarbonate: [tex]\(2.1 \, \text{g}\)[/tex]
- Total volume of [tex]\( \text{CO}_2 \)[/tex] collected: [tex]\(0.21 \, \text{dm}^3\)[/tex]
- Molar volume at RTP (Room Temperature and Pressure): [tex]\(24 \, \text{dm}^3/\text{mol}\)[/tex]
### Step-by-Step Solution:
1. Calculate the moles of [tex]\( \text{NaHCO}_3 \)[/tex] used:
[tex]\[ \text{Moles of } \text{NaHCO}_3 = \frac{\text{Initial mass of NaHCO}_3}{M_r \text{ of NaHCO}_3} \][/tex]
[tex]\[ \text{Moles of } \text{NaHCO}_3 = \frac{2.1}{84} \approx 0.025 \, \text{mol} \][/tex]
2. Stoichiometry of the reaction:
- According to the balanced equation, [tex]\(2 \text{moles of NaHCO}_3\)[/tex] produce [tex]\(1 \text{mole of Na}_2\text{CO}_3\)[/tex], [tex]\(1 \text{mole of H}_2\text{O}\)[/tex], and [tex]\(1 \text{mole of CO}_2\)[/tex].
- So, [tex]\(0.025 \, \text{moles of NaHCO}_3\)[/tex] would produce:
[tex]\[ \text{Moles of } \text{CO}_2 = \frac{0.025}{2} = 0.0125 \, \text{mol} \][/tex]
3. Expected volume of [tex]\( \text{CO}_2 \)[/tex] collected:
[tex]\[ \text{Expected volume of } \text{CO}_2 = \text{Moles of } \text{CO}_2 \times \text{Molar volume at RTP} \][/tex]
[tex]\[ \text{Expected volume of } \text{CO}_2 = 0.0125 \times 24 = 0.3 \, \text{dm}^3 \][/tex]
4. Calculate the percentage yield of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \text{Percentage yield} = \left( \frac{\text{Actual volume of } \text{CO}_2}{\text{Expected volume of } \text{CO}_2} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage yield} = \left( \frac{0.21}{0.3} \right) \times 100 = 70\% \][/tex]
5. Calculate the moles of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] produced:
- Each mole of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] corresponds to 1 mole of [tex]\( \text{CO}_2 \)[/tex] collected.
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = 0.21 \, \text{dm}^3 \div 24 \, \text{dm}^3/\text{mol} = 0.00875 \, \text{mol} \][/tex]
6. Calculate the mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] produced:
[tex]\[ \text{Mass of Na}_2\text{CO}_3 = \text{Moles of Na}_2\text{CO}_3 \times M_r \text{ of Na}_2\text{CO}_3 \][/tex]
[tex]\[ \text{Mass of Na}_2\text{CO}_3 = 0.00875 \times 106 \approx 0.9275 \, \text{g} \][/tex]
### Conclusion:
- Therefore, the mass of sodium carbonate (Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex]) produced is approximately 0.93 grams.
- The percentage yield of carbon dioxide (CO[tex]\(_2\)[/tex]) is 70%.
Thus, the correct statement is:
A. The mass of sodium carbonate produced is 0.93 g.
### Given Data:
1. Reaction:
[tex]\(2 \text{NaHCO}_3 \to \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\)[/tex]
2. Molar masses:
- [tex]\(M_r \text{of NaHCO}_3 = 84 \, \text{g/mol}\)[/tex]
- [tex]\(M_r \text{of Na}_2\text{CO}_3 = 106 \, \text{g/mol}\)[/tex]
3. Experimental data:
- Initial mass of sodium hydrogencarbonate: [tex]\(2.1 \, \text{g}\)[/tex]
- Total volume of [tex]\( \text{CO}_2 \)[/tex] collected: [tex]\(0.21 \, \text{dm}^3\)[/tex]
- Molar volume at RTP (Room Temperature and Pressure): [tex]\(24 \, \text{dm}^3/\text{mol}\)[/tex]
### Step-by-Step Solution:
1. Calculate the moles of [tex]\( \text{NaHCO}_3 \)[/tex] used:
[tex]\[ \text{Moles of } \text{NaHCO}_3 = \frac{\text{Initial mass of NaHCO}_3}{M_r \text{ of NaHCO}_3} \][/tex]
[tex]\[ \text{Moles of } \text{NaHCO}_3 = \frac{2.1}{84} \approx 0.025 \, \text{mol} \][/tex]
2. Stoichiometry of the reaction:
- According to the balanced equation, [tex]\(2 \text{moles of NaHCO}_3\)[/tex] produce [tex]\(1 \text{mole of Na}_2\text{CO}_3\)[/tex], [tex]\(1 \text{mole of H}_2\text{O}\)[/tex], and [tex]\(1 \text{mole of CO}_2\)[/tex].
- So, [tex]\(0.025 \, \text{moles of NaHCO}_3\)[/tex] would produce:
[tex]\[ \text{Moles of } \text{CO}_2 = \frac{0.025}{2} = 0.0125 \, \text{mol} \][/tex]
3. Expected volume of [tex]\( \text{CO}_2 \)[/tex] collected:
[tex]\[ \text{Expected volume of } \text{CO}_2 = \text{Moles of } \text{CO}_2 \times \text{Molar volume at RTP} \][/tex]
[tex]\[ \text{Expected volume of } \text{CO}_2 = 0.0125 \times 24 = 0.3 \, \text{dm}^3 \][/tex]
4. Calculate the percentage yield of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \text{Percentage yield} = \left( \frac{\text{Actual volume of } \text{CO}_2}{\text{Expected volume of } \text{CO}_2} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage yield} = \left( \frac{0.21}{0.3} \right) \times 100 = 70\% \][/tex]
5. Calculate the moles of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] produced:
- Each mole of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] corresponds to 1 mole of [tex]\( \text{CO}_2 \)[/tex] collected.
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = 0.21 \, \text{dm}^3 \div 24 \, \text{dm}^3/\text{mol} = 0.00875 \, \text{mol} \][/tex]
6. Calculate the mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] produced:
[tex]\[ \text{Mass of Na}_2\text{CO}_3 = \text{Moles of Na}_2\text{CO}_3 \times M_r \text{ of Na}_2\text{CO}_3 \][/tex]
[tex]\[ \text{Mass of Na}_2\text{CO}_3 = 0.00875 \times 106 \approx 0.9275 \, \text{g} \][/tex]
### Conclusion:
- Therefore, the mass of sodium carbonate (Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex]) produced is approximately 0.93 grams.
- The percentage yield of carbon dioxide (CO[tex]\(_2\)[/tex]) is 70%.
Thus, the correct statement is:
A. The mass of sodium carbonate produced is 0.93 g.
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