Question

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?

Answer 1

`"99.5 g F"_2`

Explanation:

Start by writing the balanced chemical equation that describes this reaction

`"P"_ (4(s)) + 6"F"_ (2(g)) -> 4"PF"_ (3(g))`

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.

This means that instead of

`"6 moles F"_2 " " stackrel(color(white)(acolor(blue)("at 100% yield")aaaa))(->) " " "4 moles PF"_3`

you get

`"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 color(red)(cancel(color(black)("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.