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Question
Solve the following numerical problem:
Ethane burns in oxygen according to the chemical equation:
\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
If 80 ml of ethane is burned in 300 ml of oxygen, find the composition of the resultant gaseous mixture when measured at room temperature.
Solution
\[\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}\]
Molecular weight of ethane = 30 g/mol
Molecular weight of CO2 = 44 g/mol
Molecular weight of O2 = 32 g/mol
In reaction, the number of moles of CO2 formed is twice as that of ethane.
Therefore, the volume of CO2 will be 2 × 80 = 160 ml.
In the reaction, the number of moles of O2 used is 3.5 times that of ethane.
Therefore, the volume of O2 will be 3.5 × 80 = 280 ml.
Unused oxygen is 300 – 280 ml = 20 ml.
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