Stoichiometry And Stoichiometric Calculations
A mixture of 2.0 g of H₂ and 32.0 g of O₂ is ignited. According to the reaction 2H₂ + O₂ → 2H₂O, the mass of water formed and the gas remaining after the reaction are respectively: (H = 1, O = 16)
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- stoichiometry and stoichiometric calculations
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
18 g H₂O and 16 g O₂
- Calculate Moles of Each Reactant:
- Moles of H2=22.0=1.0 mol.
- Moles of O2=3232.0=1.0 mol.
- Identify the Limiting Reagent: From the balanced equation, 2 mol H2 reacts with 1 mol O2.
- For 1.0 mol H2, we need 21.0=0.5 mol O2.
- Available O2 = 1.0 mol (excess).
- H2 is the limiting reagent.
- Calculate Product Formed:
- 2 mol H2 → 2 mol H2O, so 1.0 mol H2 → 1.0 mol H2O.
- Mass of H2O=1.0×18=18 g.
- Calculate Remaining O2:
- O2 consumed = 0.5 mol = 0.5×32=16 g.
- O2 remaining = 32−16=16 g.
- Verify by Conservation of Mass: Reactants = 2.0+32.0=34.0 g. Products = 18+16=34.0 g. Mass is conserved.
This hard 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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