Question

Consider the following reaction:

P_4(s) + 6F_{2(g) ---->} 4PF_3(g)

A, 55.0g of P₄ were mixed with 32.6g of G₂ to produce PF₃. Which reaction was the limiting reagent? Explain.

B. How many moles of PF₃ were produced i the reaction went to completion?

C. How many grams of PF₃ were collected if the reaction had 85.4% yield

Answer 1

A)

Molar mass of P4 = 123.88 g/mol

mass(P4)= 55.0 g

we have below equation to be used:

number of mol of P4,

n = mass of P4/molar mass of P4

=(55.0 g)/(123.88 g/mol)

= 0.444 mol

Molar mass of F2 = 38 g/mol

mass(F2)= 32.6 g

we have below equation to be used:

number of mol of F2,

n = mass of F2/molar mass of F2

=(32.6 g)/(38 g/mol)

= 0.8579 mol

we have the Balanced chemical equation as:

P4 + 6 F2 ---> 4 PF3 + 0

1 mol of P4 reacts with 6 mol of F2

for 0.444 mol of P4, 2.6639 mol of F2 is required

But we have 0.8579 mol of F2

so, F2 is limiting reagent

Answer: F2 is limiting reagent

B)

we will use F2 in further calculation

From balanced chemical reaction, we see that

when 6 mol of F2 reacts, 4 mol of PF3 is formed

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

= (4/6)*0.8579

= 0.5719 mol

Answer: 0.572 mol

  

C)

Molar mass of PF3 = 1*MM(P) + 3*MM(F)

= 1*30.97 + 3*19.0

= 87.97 g/mol

we have below equation to be used:

mass of PF3 = number of mol * molar mass

= 0.5719*87.97

= 50.31 g

% yield = actual mass*100/theoretical mass

85.4= actual mass*100/50.31

actual mass=42.97 g

Answer: 43.0 g