Stoichiometry
When 10.6 g of Na₂CO₃ reacts completely with excess HCl, what volume of CO₂ gas is produced at STP?
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Solution
2.24 L
- Write the balanced equation:
- Na2CO3+2HCl→2NaCl+H2O+CO2
- Find Moles of Na2CO3:
- Molar mass of Na2CO3=2(23)+12+3(16)=106 g/mol.
- Moles=106 g/mol10.6 g=0.1 mol.
- Stoichiometry:
- 1 mole Na2CO3→1 mole CO2.
- Thus, 0.1 mole of CO2 is produced.
- Calculate Volume at STP:
- Volume=moles×22.4 L
- Volume=0.1×22.4=2.24 L.
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- stoichiometry
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2.24 L
- Write the balanced equation:
- Na2CO3+2HCl→2NaCl+H2O+CO2
- Find Moles of Na2CO3:
- Molar mass of Na2CO3=2(23)+12+3(16)=106 g/mol.
- Moles=106 g/mol10.6 g=0.1 mol.
- Stoichiometry:
- 1 mole Na2CO3→1 mole CO2.
- Thus, 0.1 mole of CO2 is produced.
- Calculate Volume at STP:
- Volume=moles×22.4 L
- Volume=0.1×22.4=2.24 L.
This medium difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of stoichiometry. It appeared in the 2025 exam.
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