Question #8ebae

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"#