Question #8ebae

1 Answer
Jun 17, 2017

4848 "L"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)C2H6(g)+72O2(g)2CO2(g)+3H2O(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 = nRTPV=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"