$15 c m^{3}$ $15cm^3$ of a mixture of ethane and methane were completely oxidized by $43 c m^{3}$ $43cm^3$ of oxygen. The volumes were measured at the same temperature and pressure. What was the composition of the mixture?

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$15 c m^{3}$ $15cm^3$ of a mixture of ethane and methane were completely oxidized by $43 c m^{3}$ $43cm^3$ of oxygen. The volumes were measured at the same temperature and pressure. What was the composition of the mixture?

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Answer 1 · 2017-06-07T00:41:11

Here's what I got.

Explanation:

The idea here is that when dealing gases that are being kept under the same conditions for pressure and temperature, you can say that the mole ratios that exists between them are equivalent to volume ratios.

Start by writing the balanced chemical equations that describe these combustion reactions.

You will have

${CH}_{4 (g)} + 2 O_{2 (g)} \to {CO}_{2 (g)} + 2 H_{2} O_{(l)}$ $"CH"_ (4(g)) + color(blue)(2)"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((l))$

for the combustion of methane and

$C_{2} H_{6 (g)} + \frac{7}{2} O_{2 (g)} \to 2 {CO}_{2 (g)} + 3 H_{2} O_{(l)}$ $"C"_ 2"H"_ (6(g)) + color(purple)(7/2)"O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))$

for the combustion of ethane.

Now, notice that every mole of methane that takes part in the reaction consumes $2$ $color(blue)(2)$ moles of oxygen gas. Similarly, every mole of ethane that takes part in the reaction consumes $\frac{7}{2}$ $color(purple)(7/2)$ moles of oxygen gas.

This is equivalent to saying that

${1 cm}^{3} . methane consumes . 2 . {cm}^{3}$ $"1 cm"^3color(white)(.)"methane consumes"color(white)(.)color(blue)(2)color(white)(.)"cm"^3$ $O_{2}$ $"O"_2$

${1 cm}^{3} . ethane consumes . \frac{7}{2} . {cm}^{3}$ $"1 cm"^3color(white)(.)"ethane consumes"color(white)(.)color(purple)(7/2)color(white)(.)"cm"^3$ $O_{2}$ $"O"_2$

If you take $x$ $x$ to be the volume of methane and $y$ $y$ to be the volume of ethane present in the sample, you can say that

$xcolor(red)(cancel(color(black)("cm"^3))) + ycolor(red)(cancel(color(black)("cm"^3))) = 15 color(red)(cancel(color(black)("cm"^3)))$

which gets you

$x + y = 15$

and that

$(color(blue)(2) * x) color(red)(cancel(color(black)("cm"^3))) + (color(purple)(7/2) * y) color(red)(cancel(color(black)("cm"^3))) = 43color(red)(cancel(color(black)("cm"^3)))$

which gets you

$2x + 7/2y = 43$

Use the first equation to write

$x = 15 - y$

Plug this into the second equation to get

$2 * (15 - y) + 7/2y = 43$

Solve for $y$ to find

$4 * (15 - y) + 7y = 86$

$60 - 4y + 7y = 86$

$3y = 26 implies y = 26/3$

Consequently, you will have

$x = 15 - 26/3 = 19/3$

You can thus say that the initial mixture contained

$19/3 = 6.3color(white)(.)"cm"^3 -> "CH"_4$

$26/3 = 8.7color(white)(.)"cm"^3 ->"C"_2"H"_6$

The answers are rounded to two sig figs, the number of sig figs you have for your values.

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$15 c m^{3}$ $15cm^3$ of a mixture of ethane and methane were completely oxidized by $43 c m^{3}$ $43cm^3$ of oxygen. The volumes were measured at the same temperature and pressure. What was the composition of the mixture?

1 Answer

Explanation:

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$15 c m^{3}$ $15cm^3$ of a mixture of ethane and methane were completely oxidized by $43 c m^{3}$ $43cm^3$ of oxygen. The volumes were measured at the same temperature and pressure. What was the composition of the mixture?