
A mixture of ethane (\[{C_2}{H_6}\]) and ethene (\[{C_2}{H_4}\]) occupies 40L at 1.00 atm and 400K. The mixture reacts completely with 130g of \[{O_2}\] to produce \[C{O_2}\] and \[{H_2}O\] . Assuming ideal gas behaviour, mole fractions of \[{C_2}{H_6}\]and \[{C_2}{H_4}\]in the mixture will be:
A) \[{x_{{C_2}{H_6}}} = 0.3293,{x_{{C_2}{H_4}}} = 0.6707\]
B) \[{x_{{C_2}{H_6}}} = 0.6707,{x_{{C_2}{H_4}}} = 0.3293\]
C) \[{x_{{C_2}{H_6}}} = 0.5,{x_{{C_2}{H_4}}} = 0.5\]
D) none of these
Answer
124.5k+ views
Hint: Find the number of moles of oxygen in the given reactions. Formulate the chemical equations of the reactions to understand the different number of moles in the reactant side and the product side. Calculate the volume of oxygen required, the volume of ethene and ethane and use these values to calculate the mole fractions of each of them.
Complete step by step answer:
First, we will find the number of moles of \[{O_2}\].
In ideal gas conditions,
\[PV = nRT\]
\[ \Rightarrow 1 \times 40 = n \times 0.0821 \times 400\]
\[ \Rightarrow n = 1.218\]
One mole of ethane reacts with 3.5 moles of oxygen to give 2 moles of carbon dioxide and 2 moles of water. This can be expressed by the following chemical reaction:
\[{C_2}{H_6}(g) + \dfrac{7}{2}{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l)\]
One mole of ethene reacts with 3 moles of oxygen to give 2 moles of carbon dioxide and 2 moles of water. This can be expressed by the following chemical reaction:
\[{C_2}{H_4}(g) + 3{O_2}(g) \to 2C{O_2}(g) + 2{H_2}O(l)\]
Let the number of moles of ethene be a.
Volume of \[{O_2}\]required by ethane = \[\dfrac{7}{2}a\]
Volume of ethene = \[1.218 - a\]
Volume of \[{O_2}\]required by ethene= \[3(1.218 - a)\]
It is given that,
\[\dfrac{7}{2}a + 3(1.218 - a) = \dfrac{{130}}{{32}}\] (moles of oxygen)
\[ \Rightarrow \dfrac{7}{2}a + 3.654 - 3a = 4.0625\]
\[ \Rightarrow (1.218 - a) = (1.218 - 0.817) = 0.401\] (moles of ethene)
Mole fraction of ethane = \[\dfrac{{0.817}}{{1.218}} = 0.6707\]
Mole fraction of ethene = \[(1 - 0.6707) = 0.3293\]
Hence the correct option is B.
Note:
Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. It’s a way of expressing the concentration of a solution. Please note that mole fraction represents a fraction of molecules, and since different molecules have different masses, the mole fraction is different from the mass fraction.
Complete step by step answer:
First, we will find the number of moles of \[{O_2}\].
In ideal gas conditions,
\[PV = nRT\]
\[ \Rightarrow 1 \times 40 = n \times 0.0821 \times 400\]
\[ \Rightarrow n = 1.218\]
One mole of ethane reacts with 3.5 moles of oxygen to give 2 moles of carbon dioxide and 2 moles of water. This can be expressed by the following chemical reaction:
\[{C_2}{H_6}(g) + \dfrac{7}{2}{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l)\]
One mole of ethene reacts with 3 moles of oxygen to give 2 moles of carbon dioxide and 2 moles of water. This can be expressed by the following chemical reaction:
\[{C_2}{H_4}(g) + 3{O_2}(g) \to 2C{O_2}(g) + 2{H_2}O(l)\]
Let the number of moles of ethene be a.
Volume of \[{O_2}\]required by ethane = \[\dfrac{7}{2}a\]
Volume of ethene = \[1.218 - a\]
Volume of \[{O_2}\]required by ethene= \[3(1.218 - a)\]
It is given that,
\[\dfrac{7}{2}a + 3(1.218 - a) = \dfrac{{130}}{{32}}\] (moles of oxygen)
\[ \Rightarrow \dfrac{7}{2}a + 3.654 - 3a = 4.0625\]
\[ \Rightarrow (1.218 - a) = (1.218 - 0.817) = 0.401\] (moles of ethene)
Mole fraction of ethane = \[\dfrac{{0.817}}{{1.218}} = 0.6707\]
Mole fraction of ethene = \[(1 - 0.6707) = 0.3293\]
Hence the correct option is B.
Note:
Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. It’s a way of expressing the concentration of a solution. Please note that mole fraction represents a fraction of molecules, and since different molecules have different masses, the mole fraction is different from the mass fraction.
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