Question

Phosphorus and chlorine react as follows: P4(s) + Cl2(g) => PCl3(l). Balance it! Assume 135 g P4(s) react with 333 g of Cl2(g).

a) What is the limiting reactant?

b) What is the reactant in excess and the amount in excess?

c) What mass of phosphorus trichloride can be formed from the reaction?

Answer 1

a)

mass of P4 = 135.0 g

molar mass of P4 = 123.9 g/mol

mol of P4 = (mass)/(molar mass)

= 135.0/123.9

= 1.1 mol

mass of Cl2 = 333.0 g

molar mass of Cl2 = 70.9 g/mol

mol of Cl2 = (mass)/(molar mass)

= 333.0/70.9

= 4.7 mol

Balanced chemical equation is:

1P4 + 6Cl2 ---> 4PCl3

1 mol of P4 reacts with 6 mol of Cl2

for 1.1 mol of P4 6.6 mol of Cl2 is required

But we have 4.7 mol of Cl2

Cl2 is limiting reagent

b)

we will use Cl2 in further calculation

According to balanced equation

mol of P4 reacted = (1/6)* moles of Cl2

= (1/6)*4.7

= 0.783 mol

Mol of P4 remaining = 1.1 mol - 0.783 mol = 0.317 mol

Mass of P4 excess = mol of P4 remaining * molar mass of P4

= 0.317*123.9

= 39.3 g

c)

According to balanced equation

mol of PCl3 formed = (4/6)* moles of Cl2

= (4/6)*4.7

= 3.13 mol

mass of PCl3 = number of mol * molar mass

= 3.13*137.3

= 430 g