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?
1 Answer
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
This represents the reaction's theoretical yield, i.e. what you get for a reaction that has a
In your case, the reaction is said to have a
This is equivalent to saying that for every
#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.