Answer:
44.28 grams.
Explanation:
Let us write the balanced reaction:
[tex]P_{4}+6F_{2}-->4PF_{3}[/tex]
As per balanced equation, six moles of fluorine gas will give four moles of PF₃.
The mass of PF₃ required = 120 g
The molar mass of PF₃ = 88g/mol
Moles of PF₃ required =[tex]\frac{mass}{molarmass}=\frac{120}{88}=1.364mol[/tex]
The moles of fluorine gas required = [tex]\frac{4X1.364}{6}=0.91[/tex]
the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g
Now this much mass will be required if the reaction is of 100% yield
But as given that the yield of reaction is only 78.1%
The mass of fluorine required = [tex]\frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g[/tex]
