# For the reaction "PCl"_3(g) + "Cl"_2(g) -> "PCl"_5(g), the percent yield was 88.2% for a certain experiment. If "77.4 g" of "PCl"_3 was consumed in the presence of xs "Cl"_2(g), what would be the actual yield of "PCl"_5(g) for this experiment?

Jan 14, 2017

${\text{103.52 g PCl}}_{5}$

You are given the reaction:

${\text{PCl"_3(g) + "Cl"_2(g) -> "PCl}}_{5} \left(g\right)$

Since you are given the percent yield of 88.2%, it means that the $\text{Actual Yield}$ is

$0.882 \times \text{Theoretical Yield}$
(which is what you're asked for!),

since

%"Yield" = ("Actual Yield")/("Theoretical Yield")xx100%.

So, you need to calculate the theoretical yield from the mass of the reactant ${\text{PCl}}_{3}$, knowing already that it is the limiting reagent (since ${\text{Cl}}_{2}$ was stated to be in excess).

The limiting reagent simply gets used up first in the reaction, and therefore limits the yield of the reaction (it controls how much you can theoretically make of the product).

Since $\text{1 mol}$ of something is $\text{1 mol}$ of anything, this is a matter of converting like so:

"mass of limiting reagent (PCl"_3)

-> "mols of limiting reagent (PCl"_3)

-> "mols of product (PCl"_5)

-> "mass of product (PCl"_5)

Whatever's not in parentheses in the procedure above is the general process for limiting reagent calculations.

The result is your theoretical yield, which you can use to get the actual yield, which is what was asked for.

$77.4 \cancel{{\text{g PCl"_3) xx (cancel("1 mol PCl"_3))/(137.33 cancel("g PCl"_3)) xx cancel("1 mol PCl"_5)/cancel("1 mol PCl"_3) xx ("208.24 g PCl"_5)/cancel("1 mol PCl}}_{5}}$

$=$ ${\text{117.37 g PCl}}_{5}$

But since this is your theoretical yield, the actual yield is:

$\textcolor{b l u e}{{\text{Actual Yield") = "117.37 g" xx "0.882" = color(blue)("103.52 g PCl}}_{5}}$