Question 6cd3c

Apr 17, 2015

The theoretical yield of sulfur trioxide, or $S {O}_{3}$, will be 285.0 mL, and the percent yield will be 61.16%.

$\textcolor{red}{2} S {O}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to \textcolor{red}{2} S {O}_{3 \left(g\right)}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between sulfur dioxide, $S {O}_{2}$, and oxygen; this will be very important when comparing the number of moles of each gas that react, since it will tell you if one of the two compounds will act as a limiting reagent.

Your tool of choice will be the ideal gas law equation. Start by determining the number of moles of sulfur dioxide and oxygen that you start with

$P V = n r t \implies n = \frac{P V}{R T}$

n_(O_2) = (48.5/760cancel("atm") * 159 * 10^(-3)cancel("L"))/(0.082 (cancel("L") * cancel("atm"))/("mol" * cancel("K")) * 330cancel("K")) = "0.0003750 moles " ${O}_{2}$

n_(SO_2) = (48.5/760cancel("atm") * 284.9 * 10^(-3)cancel("L"))/(0.082 (cancel("L") * cancel("atm"))/("mol" * cancel("K")) * 330cancel("K")) = "0.0006719 moles " $S {O}_{2}$

According to the mole ratio, that many moles of oxygen would have required

0.0003750cancel("moles "O_2) * (color(red)(2)"moles "SO_2)/(1cancel("mole "SO_2)) = "0.000750 moles " $S {O}_{2}$

Since you've got fewer moles than required, sulfur dioxide will act as a limiting reagent, i.e. it will determine how many moles of oxygen will actually react

0.0006719cancel("moles "SO_2) * ("1 mole "O_2)/(color(red)(2)cancel("moles "SO_2)) = "0.0003360 moles "O_2

The number of moles of sulfur trioxide this reaction will produce is

0.0006719cancel("moles "SO_2) * ("1 mole "SO_3)/(1cancel("mole "SO_2)) = "0.0006719 moles "SO_3

This many moles will produce a volume of

$V = \frac{n R T}{P} = \left(0.0006719 \cancel{\text{moles") * 0.082(("L" * cancel("atm"))/(cancel("mol") * cancel("K")) * 330cancel("K")))/(48.5/760cancel("atm}}\right)$

V = "0.285 L" = color(green)("285.0 mL")

This represents the theoretical yield of $S {O}_{3}$ $\to$ this is how much sulfur trioxide your reaction would produce if all (100%) of the sulfur dioxide reacts.

SInce your reaction produced 174.3 mL, the percent yield of the reaction will be smaller than 100%. Calculate percent yield by

$\text{% yield" = "actual yield"/"theoretical yield} \cdot 100$

"% yield" = 174.3/285.0 * 100 = color(green)("61.16%")#