What mass of f2 is needed to produce 120. g of pf3 if the reaction has a 78.1% yield?

write the equation for the reaction

that is 6 F2 + P4 = 4 PF3

find the theoretical mass that is

let the theoretical yield be represented by y

theoretical yield = 78.1/100 = 120/y

y = 153.6 grams

find the number of moles of PF3

moles = mass/molar mass

= 153.6/87.97 = 1.746 moles

by use of mole ratio between F2 : PF3 which is 6:4 the moles of F2 is therefore = 1.746 x 6/4 = 2.62 moles

mass = moles x molar mass

= 1.746 moles x38 g/mol = 99.6 grams