Question

Consider the following unbalanced reaction. P4(s) + F2(g) PF3(g) How many grams of F2 are needed to produce 186 g of PF3 if the reaction has a 76.3% yield?

Answer 1

The balanced chemical reaction is:

P4(s) + 6F2(g) --> 4PF3(g)

actual mass = 186 g

use:

% yield = actual mass *100 / theoretical mass

76.3 = 186*100 / m

m = 243.8 g

Molar mass of PF3,

MM = 1*MM(P) + 3*MM(F)

= 1*30.97 + 3*19.0

= 87.97 g/mol

mass(PF3)= 243.8 g

number of mol of PF3,

n = mass of PF3/molar mass of PF3

=(243.8 g)/(87.97 g/mol)

= 2.771 mol

from balanced reaction,

moles of F2 required = (6/4)*moles of PF3 formed

= (6/4)*2.771 mole

= 4.16 moles

Molar mass of F2 = 38 g/mol

mass of F2,

m = number of mol * molar mass

= 4.16 mol * 38 g/mol

= 158 g

Answer: 158 g