# Question 8ebae

Jun 17, 2017

$48$ $\text{L}$

#### Explanation:

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

$\text{C"_2"H"_6(g) + 7/2"O"_2(g) rarr 2"CO"_2(g) + 3"H"_2"O} \left(g\right)$

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:

$P V = n R T$

Our known values are

• P = 200cancel("kPa")((1"atm")/(101.325cancel("kPa"))) = 1.97 $\text{atm}$

• $V = 16$ $\text{L}$

• $R = 0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$ (universal gas constant)

• $T = {5000}^{\text{o""C}} + 273 = 5273$ $\text{K}$

I converted the units where necessary to make sure I had the units $\text{atm}$, $\text{L}$, and $\text{K}$.

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, $\text{H"_2"O} \left(g\right)$:

0.0730cancel("mol C"_2"H"_6)((3"mol H"_2"O")/(1cancel("mol C"_2"H"_6))) = 0.219 $\text{mol H"_2"O}$

Now, let's again use the ideal-gas equation to solve for the volume of the $\text{H"_2"O} \left(g\right)$, using the same conditions as before.

Known values:

• $P = 1.97$ $\text{atm}$

• $n = 0.219$ $\text{mol}$

• $R = 0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$

• $T = 5273$ $\text{K}$

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, $16$ $\text{L}$. This is a result of the coefficients of the chemical equation, with $\text{H"_2"O} \left(g\right)$ having a coefficient of $3$, and ${\text{C"_2"H}}_{6} \left(g\right)$ with a coefficient of $1$.

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

${V}_{\text{H"_2"O}} = 16$ "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"#