Stoichiometry And Stoichiometric Calculations
When 5.3 g of Na₂CO₃ reacts with excess HCl according to the equation Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂, the volume of CO₂ produced at STP is approximately: (Na = 23, C = 12, O = 16)
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- stoichiometry and stoichiometric calculations
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.12 L
- Calculate Molar Mass of Na₂CO₃:
- M=2(23)+12+3(16)=46+12+48=106 g/mol.
- Find Moles of Na₂CO₃:
- n=1065.3=0.05 mol.
- Use Stoichiometry: From the balanced equation, 1 mol Na₂CO₃ produces 1 mol CO₂.
- Therefore, moles of CO₂ = 0.05 mol.
- Calculate Volume at STP:
- At STP, 1 mol of any ideal gas occupies 22.4 L.
- V=0.05×22.4=1.12 L.
- Sanity Check: 5.3 g is exactly 1/20 of the molar mass of Na₂CO₃, so we expect 1/20 of 22.4 L, confirming 1.12 L.
This medium difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of stoichiometry and stoichiometric calculations. It appeared in the 2025 exam.
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