Question

An ideal gas mixture of C\(_2\)H\(_6\) and C\(_2\)H\(_4\) occupies a volume of 28 L at 1 atm and 273 K. This mixture reacts completely with 128 g of O\(_2\) to produce CO\(_2\) and H\(_2\)O(\(l\)). What is the mole fraction of C\(_2\)H\(_4\) in the mixture ?

• A. \(0.4 \)
• B. \(0.8 \)
• C. \(0.5 \)
• D. \(0.6 \)

Hint:

When solving problems involving gas mixtures and chemical reactions: 1. Ideal Gas Law: Use \(PV=nRT\) to find the total moles of the gas mixture if pressure, volume, and temperature are given. 2. Balanced Equations: Write balanced chemical equations for all reactions involved, paying attention to the stoichiometry (mole ratios). 3. System of Equations: If there are multiple components and total quantities (like total moles or total reactant consumed), set up a system of simultaneous equations. 4. Mole Fraction: Mole fraction of a component = (moles of component) / (total moles of mixture).

Answer 1

The Correct Option is D

Step 1: Calculate the total moles of the gas mixture using the Ideal Gas Law.
Given:
Volume (\(V\)) = 28 L
Pressure (\(P\)) = 1 atm
Temperature (\(T\)) = 273 K
Gas Constant (\(R\)) = 0.0821 L atm mol\(^{-1}\) K\(^{-1}\) (Note: At 273 K and 1 atm, this is STP conditions, where 1 mole of an ideal gas occupies 22.4 L).
Using the Ideal Gas Law, \(PV = nRT\):
\[ n_{total} = \frac{PV}{RT} = \frac{(1 \operatorname{atm})(28 \operatorname{L})}{(0.0821 \operatorname{L} \operatorname{atm} \operatorname{mol}^{-1} \operatorname{K}^{-1})(273 \operatorname{K})} \] Alternatively, using the molar volume at STP: \[ n_{total} = \frac{28 \operatorname{L}}{22.4 \operatorname{L/mol}} = 1.25 \operatorname{mol} \] So, the total moles of the gas mixture (C\(_2\)H\(_6\) + C\(_2\)H\(_4\)) is 1.25 mol. 
Step 2: Write balanced chemical equations for the complete combustion of C\(_2\)H\(_6\) and C\(_2\)H\(_4\).
Combustion of ethane (C\(_2\)H\(_6\)):
\[ \operatorname{C}_2\operatorname{H}_6(\operatorname{g}) + \frac{7}{2}\operatorname{O}_2(\operatorname{g}) \rightarrow 2\operatorname{CO}_2(\operatorname{g}) + 3\operatorname{H}_2\operatorname{O}(\operatorname{l}) \] This means 1 mole of C\(_2\)H\(_6\) requires 3.5 moles of O\(_2\).
Combustion of ethene (C\(_2\)H\(_4\)):
\[ \operatorname{C}_2\operatorname{H}_4(\operatorname{g}) + 3\operatorname{O}_2(\operatorname{g}) \rightarrow 2\operatorname{CO}_2(\operatorname{g}) + 2\operatorname{H}_2\operatorname{O}(\operatorname{l}) \] This means 1 mole of C\(_2\)H\(_4\) requires 3 moles of O\(_2\). 
Step 3: Calculate the moles of O\(_2\) consumed.
Given mass of O\(_2\) = 128 g.
Molar mass of O\(_2\) = \(2 \times 16.00 = 32.00 \operatorname{g/mol}\).
\[ \text{Moles of O}_2 = \frac{128 \operatorname{g}}{32 \operatorname{g/mol}} = 4.00 \operatorname{mol} \] 
Step 4: Set up a system of equations based on moles.
Let \(x\) be the moles of C\(_2\)H\(_6\) and \(y\) be the moles of C\(_2\)H\(_4\) in the mixture.
From Step 1 (total moles of mixture): \[ x + y = 1.25 \quad \text{(Equation 1)} \] From Step 2 and 3 (total moles of O\(_2\) consumed):
Moles of O\(_2\) from C\(_2\)H\(_6\) = \(3.5x\)
Moles of O\(_2\) from C\(_2\)H\(_4\) = \(3y\)
Total moles of O\(_2\) = \(3.5x + 3y = 4.00 \quad \text{(Equation 2)}\)
Step 5: Solve the system of equations for \(y\).
From Equation 1, \(x = 1.25 - y\). Substitute this into Equation 2: \[ 3.5(1.25 - y) + 3y = 4.00 \] \[ 4.375 - 3.5y + 3y = 4.00 \] \[ 4.375 - 0.5y = 4.00 \] \[ 0.5y = 4.375 - 4.00 \] \[ 0.5y = 0.375 \] \[ y = \frac{0.375}{0.5} = 0.75 \operatorname{mol} \] So, the moles of C\(_2\)H\(_4\) is 0.75 mol. 
Step 6: Calculate the mole fraction of C\(_2\)H\(_4\).
Mole fraction of C\(_2\)H\(_4\), \(\chi_{C_2H_4} = \frac{\text{moles of C}_2H_4}{\text{total moles of mixture}}\).
\[ \chi_{C_2H_4} = \frac{y}{n_{total}} = \frac{0.75 \operatorname{mol}}{1.25 \operatorname{mol}} \] \[ \chi_{C_2H_4} = \frac{0.75}{1.25} = \frac{75}{125} \] Divide both numerator and denominator by 25: \[ \chi_{C_2H_4} = \frac{3}{5} = 0.6 \] The final answer is \(\boxed{0.6}\).