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Difficulty: MediumGay-Lussac's Law of Combining Volumes and Avogadro's Law

A mixture of 25 cm325\text{ cm}^3 of methane (CH4CH_4) and 60 cm360\text{ cm}^3 of oxygen (O2O_2) is sparked at constant temperature and pressure. Assuming water vapor condenses to liquid upon cooling to room temperature, what is the total volume of residual gas remaining in cm3\text{cm}^3?

Answer: 35 cm³

Answer

The total volume of residual gas remaining after cooling to room temperature is 35 cm335\text{ cm}^3.
According to Gay-Lussac's law of combining volumes, gases react in simple numerical ratios equal to their stoichiometric coefficients at constant temperature and pressure. For the equation CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l), 25 cm325\text{ cm}^3 of methane reacts completely with 50 cm350\text{ cm}^3 of oxygen to produce 25 cm325\text{ cm}^3 of CO2CO_2 gas. Since 60 cm360\text{ cm}^3 of oxygen was initially present, 10 cm310\text{ cm}^3 of oxygen remains unreacted. Liquid water occupies negligible volume compared to gases. Therefore, the total residual gaseous volume is the sum of unreacted oxygen and produced carbon(IV) oxide: 10 cm3+25 cm3=35 cm310\text{ cm}^3 + 25\text{ cm}^3 = 35\text{ cm}^3.

Step-by-Step Solution

1
Write the balanced chemical equation for the combustion of methane gas.
CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)
Gay-Lussac's law of combining volumes applies directly to the stoichiometric mole ratios of gaseous reactants and products.
2
Determine the volume of oxygen consumed and identify the excess reactant.
25 cm325\text{ cm}^3 of CH4CH_4 consumes 50 cm350\text{ cm}^3 of O2O_2, leaving 10 cm310\text{ cm}^3 of unreacted O2O_2.
The reaction stoichiometry requires 2 volumes of oxygen per 1 volume of methane.
3
Calculate the volume of gaseous carbon(IV) oxide produced.
Volume of CO2(g)=25 cm3CO_2(g) = 25\text{ cm}^3.
1 volume of methane produces 1 volume of carbon(IV) oxide gas.
4
Sum the volumes of all remaining gaseous species.
Total residual volume = 10 cm3 (excess O2)+25 cm3 (formed CO2)=35 cm310\text{ cm}^3\text{ (excess } O_2) + 25\text{ cm}^3\text{ (formed } CO_2) = 35\text{ cm}^3.
Water formed is liquid at room temperature and contributes negligibly to gaseous volume.

Key Concept

Gay-Lussac's Law of Combining Volumes and Stoichiometry of Gas Reactions
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Topics

Gay-Lussac's Law of Combining Volumes and Avogadro's Law

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