Question

Question #8ebae

Answer 1

`48` `"L"`

Explanation:

I'm not given a reaction, but I'm going to assume that the ethane undergoes combustion, and thus the equation would be

`"C"_2"H"_6(g) + 7/2"O"_2(g) rarr 2"CO"_2(g) + 3"H"_2"O"(g)`

What we can do first in solving this equation is using the given conditions to calculate the number of moles of ethane used up, using the ideal-gas equation:

`PV = nRT`

Our known values are

• `P = 200cancel("kPa")((1"atm")/(101.325cancel("kPa"))) = 1.97` `"atm"`

• `V = 16` `"L"`

• `R = 0.082057("L"·"atm")/("mol"·"K")` (universal gas constant)

• `T = 5000^"o""C" + 273 = 5273` `"K"`

I converted the units where necessary to make sure I had the units `"atm"`, `"L"`, and `"K"`.

Let's now plug in these variables, and solve for `n`, the number of moles:

`n = (PV)/(RT) = ((1.97cancel("atm"))(16cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(5273cancel("K"))) = color(red)(0.0728` `color(red)("mol C"_2"H"_6`

Now, we can use this value and the coefficients in the chemical equation to find the relative number of moles of water vapor, `"H"_2"O"(g)`:

`0.0730cancel("mol C"_2"H"_6)((3"mol H"_2"O")/(1cancel("mol C"_2"H"_6))) = 0.219` `"mol H"_2"O"`

Now, let's again use the ideal-gas equation to solve for the volume of the `"H"_2"O"(g)`, using the same conditions as before.

Known values:

• `P = 1.97` `"atm"`

• `n = 0.219` `"mol"`

• `R = 0.082057("L"·"atm")/("mol"·"K")`

• `T = 5273` `"K"`

Plugging them in and solving for the volume, `V`:

`V = (nRT)/P = ((0.219cancel("mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(5273cancel("K")))/(1.97cancel("atm")) = color(blue)(48` `color(blue)("L"`

Notice that the volume is simply triple that of ethane, `16` `"L"`. This is a result of the coefficients of the chemical equation, with `"H"_2"O"(g)` having a coefficient of `3`, and `"C"_2"H"_6(g)` with a coefficient of `1`.

We therefore could also have solved this using just the volume and the coefficients:

`V_ ("H"_2"O") = 16` `"L C"_2"H"_6((3 "(coefficient H"_2"O)")/(1 "(coefficient C"_2"H"_6")")) = color(blue)(48` `color(blue)("L H"_2"O"`