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Question
40 cm3 of methane (CH4) is reacted with 60 cm3 of oxygen.
The equation for the reaction is given below:
\[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]
All volumes are measured at room temperature.
What is the total volume of the gases remaining at the end of the reaction?
Options
60 cm3
40 cm3
45 cm3
50 cm3
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Solution
40 cm3
Explanation:
\[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]
According to Gay Lussac’s Law, one volume of methane combines with two volumes of oxygen to produce one volume of carbon dioxide. As water (H2O) is a liquid, it does not contribute to the overall volume of gases.
Therefore, 40 cm3 of CH4 would require 2 × 40 cm3 = 80 cm3 of O2.
Since only 60 cm3 of O2 is available, 60 cm3 of O2 reacts with `1/2 xx 60` cm3 = 30 cm3 of CH4 is used. Now, the volume of remaining CH4 will be 40 cm3 − 30 cm3 = 10 cm3.
According to equation, 60 cm3 of O2 will produce `1/2 xx 60` cm3 = 30 cm3 of CO2.
Therefore, the total volume of gases remaining is the sum of the volume of CH4 remaining and the volume of CO2 produced.
So, total volume = 10 cm3 + 30 cm3 = 40 cm3.
