22.4 dm3 of volume of each H2 and O2 are sparked to produce water vapours (considered as ideal gas) on completion of reaction. What is the decrease in the volume of the vessel?
Correct answer: D. 11.2 dm3
- A. 44 dm3
- B. 22.4 dm3
- C. 33.6 dm3
- D. 11.2 dm3
Explanation
In the reaction, 2 moles of H2 react with 1 mole of O2 to form water vapor. Initially, there are 44.8 dm3 of gas (22.4 dm3 of each gas). According to the stoichiometry, 22.4 dm3 of H2 reacts with only 11.2 dm3 of O2, forming 22.4 dm3 of water vapor. The remaining 11.2 dm3 of O2 adds to the water vapor for a total final volume of 33.6 dm3. The initial volume was 44.8 dm3, so the decrease in the volume is 44.8 - 33.6 = 11.2 dm3. The other options misunderstand either the stoichiometry or the ideal gas volume concept.
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