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Chapter 5: Problem 166

A \(9.780-\mathrm{g}\) gaseous mixture contains ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Complete combustion to form carbon dioxide and water requires 1.120 mole of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Short Answer

Expert verified
The mass percent of ethane in the original gaseous mixture is approximately 21.14%.

Step by step solution

01

Write the balanced combustion equations for ethane and propane

The balanced combustion equations for ethane and propane are \(C_2H_6 + \dfrac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O\) \(C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O\) These equations tell us that 1 mole of ethane requires \(\dfrac{7}{2}\) moles of oxygen for combustion, while 1 mole of propane requires 5 moles of oxygen.
02

Set up a system of equations to find the moles of ethane and propane

Let x represent the moles of ethane and y represent the moles of propane. Then, we have the following system of equations: \(x\left(\dfrac{7}{2}\right) + 5y = 1.120\) moles of oxygen (from the balanced equations) \(30x + 44y = 9.780\) g (total mass of the ethane and propane mixture)
03

Solve the system of equations

To solve the system of equations, we can multiply the first equation by -30 and then add the two equations to eliminate x: \(-15x - 30y = -3.360\) \(-30x - 14y = -9.780\) Now, add the equations: \(-44y = -13.140\) Divide by -44 to solve for y: \(y = 0.29886\) moles of propane Now, substitute y back into the first equation to solve for x: \(x\left(\dfrac{7}{2}\right) + 5(0.29886) = 1.120\) \(x = 0.11762\) moles of ethane
04

Convert moles to mass and find the mass percent of ethane

Now that we have the moles of ethane and propane, we can convert them to mass and find the mass percent of ethane. Mass of ethane = moles of ethane × molar mass of ethane \(Mass_{ethane} = 0.11762 \times 30 = 3.5286\;\mathrm{g}\) Mass of propane = moles of propane × molar mass of propane \(Mass_{propane} = 0.29886 \times 44 = 13.14984\;\mathrm{g}\) Now, we can find the mass percent of ethane: Mass percent of ethane = \(\dfrac{Mass_{ethane}}{Mass_{ethane} + Mass_{propane}} \times 100\) Mass percent of ethane = \(\dfrac{3.5286}{3.5286 + 13.14984} \times 100 \approx 21.14\%\) Therefore, the mass percent of ethane in the original mixture is approximately 21.14%.

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Most popular questions from this chapter

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