Question

What mass of F2 is needed to produce 120.0 g of PF3, as shown, if the reaction has
a 78.1% yield?
P4 (s) + 6 F2 (g) = 4 PF3​

Answer 1

Answer:

99.6 grams

Explanation:

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