Question

A mixture of 20.6g of phosphorus (P) and 79.4 chlorine (Cl2) reacts completely to form phosphorus trichloride (PCl3) and phosphorus pentachloride (PCl5).

(a) true or false: the bond angels of PCl3 and PCl5 are identical.

(b) what mass of PCl3 forms under the conditions described above?

(c) an analysis of a phosphorus and chlorine containing compound found that it was 30.40% phosphorus. is it one of the two products formed in this reaction? why pr why not?

Answer 1

Given that 20.6 g of phosphorus (P) and 79.4 g of chlorine (Cl2) reacts completely to form phosphorus trichloride (PCl3) and phosphorus pentachloride (PCl5).

a) False

   PCl₃ = pyramidal shape =   < 109^o

PCl₅ = Trigonal bipyramidal shape = 90^o, 120^o

Hence,

bond angels of PCl3 and PCl5 are not identical.

b)    2P + 4Cl2 -------------> PCl3 + PCl5

    Moles of P = mass / molar mass in balanced equation = 20.6 g/ 2x 31 g/mol = 0.332 mol

       Moles of Cl2 = mass / molar mass in balanced equation = 79.4 g/ 4 x 35.5 g/mol = 0.559 mol

Moles of P are less than moles of Cl2.

Hence, P is the limiting reagent.

Yield of PCl₃ is calculated based on P.

2P         + 4 Cl2 ------------->    PCl3 +   PCl5

2 mol                                       1 mol

2 x 31 g = 62 g                                137.5 g

20.6 g                                                ?

  

                     ? = (20.6 g/ 62 g) x 137.5 g PCl3

                        = 45.68 g PCl3

Therefore,

mass of PCl3 formed = 45.68 g

c) PCl3:

   Molar mass of PCl3 = 137.5 g/mol

Molar mass of P = 31 g

Mass % of P in PCl3 = P/PCl3 = 31 g / 137.5 g x 100 = 22.5 %

PCl5:

   Molar mass of PCl5 = 208.5 g/mol

   Molar mass of P = 31 g

Mass % of P in PCl5 = P/PCl3 = 31 g / 208.5 g x 100 = 14.8 %

.....................................

Given Compound has 30.40% phosphorus.

Therefore, it is neither PCl3 nor PCl5.