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A mixture of ethane and ethene occupies $40$ litre at $1$ atm and $400$ K. The mixture reacts completely with $130$ g of Oxygen​ to produce Carbon dioxide​ and water. Assuming ideal gas behaviour, calculate the mole fractions of ethane​ and ethene​ in the mixture.

Last updated date: 15th Jun 2024
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Hint:The mole fraction of a solute is equal to the ratio of the number of moles of that solute to the total number of moles of solute and solvent in the solution. Since it's a ratio of moles to moles, the mole fraction is a dimensionless number, and of course, it is always less than one.
For a solution, the mole fraction of solute A is $ = $ (moles of A) $ \div $ (total moles), and the mole fraction of the solvent $ = $ (moles of solvent) $ \div $ (total moles).

Complete step by step answer:
Applying the ideal gas equation below:
The volume of the mixture at NTP is
$\dfrac{{40 \times 1}}{{400}} \times \dfrac{{273}}{1} = 27.3$
Let the volume of ethane be = x litre
The volume of ethane = \[\left( {273 - x} \right)\]litre
Balanced equation is as follows:
\[{C_2}{H_6} + 7/2{O_2} \to 2C{O_2} + 3{H_2}\]
\[{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\]
The total volume of oxygen required for the complete combustion of the mixture is:
\[\left[ {\dfrac{7}{2}x + \left( {27.3 - x} \right) \times 3} \right]\]Litre
or \[(\dfrac{7}{2}x + \left( {27.3 - x} \right) \times 3) \div 2\]litre
Mass of oxygen $ = \left[ {\dfrac{7}{2}x + \left( {27.3 - x} \right) \times 3} \right] \times \dfrac{{32}}{{22.4}}$
\[130 = \left( {x + 163.8} \right) \times 22.416\]
∴\[x = 18.2\]
Hence, mole fraction of ethane =\[\dfrac{{18.2}}{{27.3}} \times 100 = 66.66\]
Mole fraction of ethane \[ = 33.34\]

-Ethane is an alkane and per se is saturated, meaning it's all single bonds. Ethene is an alkene and intrinsically unsaturated, meaning it's one carbon to carbon covalent bond. so that they are often identified by their displayed formulae, as ethene has the covalent bond functional group.
-Chemically, they'll be distinguished most easily by the reaction with bromine water, which is orange/red in colour. Add some drops to ethene and bromine water is decolourised as $1,2 - $ dibromoethane, a saturated compound. There's no visible sign of reaction when bromine is added to an alkane.