# White phosphorus, "P"_4, is an allotrope of phosphorus that reacts with fluorine gas to form gaseous phosphorus trifluoride. What is the mass of fluorine gas needed to produce "120. g" of phosphorus trifluoride if the reaction has a 78.1% yield?

Aug 22, 2017

${\text{99.5 g F}}_{2}$

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction

${\text{P"_ (4(s)) + 6"F"_ (2(g)) -> 4"PF}}_{3 \left(g\right)}$

Notice that for every $6$ moles of fluorine gas that take part in the reaction, the reaction produces $4$ moles of phosphorus trifluoride.

This represents the reaction's theoretical yield, i.e. what you get for a reaction that has a 100% yield.

In your case, the reaction is said to have a 78.1% percent yield. This means that for every $100$ moles of phosphorus trifluoride that the reaction could theoretically produce, you only get $78.1$ moles.

This is equivalent to saying that for every $6$ moles of fluorine gas that the reaction consumes, you only get

4 color(red)(cancel(color(black)("moles PF"_3))) * "78.1 moles PF"_3/(100color(red)(cancel(color(black)("moles PF"_3)))) = "3.124 moles PF"_3

This represents the reaction's actual yield, i.e. what you actually get when you perform the reaction.

${\text{6 moles F"_2 " " stackrel(color(white)(acolor(blue)("at 100% yield")aaaa))(->) " " "4 moles PF}}_{3}$

you get

${\text{6 moles F"_2 " " stackrel(color(white)(acolor(blue)("at 78.1% yield")aaaa))(->) " " "3.124 moles PF}}_{3}$

So, convert the mass of phosphorus trifluoride to moles by using the compound's molar mass

120. color(red)(cancel(color(black)("g"))) * "1 mole PF"_3/(87.97color(red)(cancel(color(black)("g")))) = "1.363 moles PF"_3

This means that the reaction must have consumed

1.363 color(red)(cancel(color(black)("moles PF"_3))) * "6 moles F"_2/(3.124 color(red)(cancel(color(black)("moles PF"_3)))) = "2.618 moles F"_2

To convert this to grams, use the molar mass of fluorine gas

$2.618 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles F"_2))) * "37.997 g"/(1color(red)(cancel(color(black)("mole F"_2)))) = color(darkgreen)(ul(color(black)("99.5 g F}}_{2}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of phosphorus trifluoride produced by the reaction.