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Chapter 3: Problem 92

A 0.732 g mixture of methane, \(\mathrm{CH}_{4}\), and ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) is burned, yielding \(2.064 \mathrm{g} \mathrm{CO}_{2} .\) What is the percent composition of this mixture (a) by mass; (b) on a mole basis?

Short Answer

Expert verified
Using step 1 to 3, x (Moles of methane) and y (Moles of ethane) can be calculated. Once these are known, the composition by mass and mole basis can be calculated using formulas described in step 4.

Step by step solution

01

Reaction Stoichiometry

The combustion reactions for Methane and Ethane are:\[\mathrm{CH_{4 (g)} + 2O_{2 (g)} -> CO_{2 (g)} + 2H_{2O (g)}}\]and\[\mathrm{C_{2}H_{6 (g)} + 7/2O_{2 (g)} -> 2CO_{2 (g)} + 3H_{2O (g)}}\]which implies that 1 mole of Methane or 1 mole of Ethane will produce 1 or 2 moles of Carbon Dioxide respectively.
02

Formation of Equations

Now consider the initial moles of Methane as x and of Ethane as y. Since 1 mole of CH4 and 1 mole of C2H6 will produce 1 and 2 moles of CO2 respectively, the moles of CO2 produced are x + 2y. Also, as the mass of the gas mixture is known (0.732 g), can formulate the following equations:\[\mathrm{16.04x + 30.07y = 0.732} (equation 1)\]and\[\mathrm{x + 2y = mass of CO_{2} produced / 44.01} (equation 2)\]since 1 mole of CO2 has a mass of 44.01 g.
03

Solving the Equations

Substitute the given mass of CO2 produced (2.064 g) in equation 2 to get the moles of CO2 and then solve the 2 equations simultaneously to get the values of x and y. This will give the initial moles of CH4 and C2H6.
04

Calculation of % Composition

Once the x and y values are known, we can use the following formulas to determine the percent composition by mass and on a mole basis: \[\%composition (mass basis) = (mass component / mass total) * 100\]\[\%composition (mole basis) = (moles component / moles total) * 100\]

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