Question

Elemental phosphorus reacts with chlorine gas according to the equation

P4(s)+6Cl2(g)→4PCl3(l)
A reaction mixture initially contains 45.57 g P4 and 130.2 g Cl2.

Once the reaction has reached completion, what mass (in g) of the excess reactant is left?
(Express the mass in grams to three significant figures.)

Answer 1

Mass (in g) of the excess reactant (P₄) left = 7.626 g

Further explanation

Given

Reaction

P4(s)+6Cl2(g)→4PCl3(l)

45.57 g P4 and 130.2 g Cl2

Required

mass (in g) of the excess reactant left

Solution

mol P₄ (MW=124 g/mol) :

45.57 : 124 = 0.3675

mol Cl₂(MW=71 g/mol) :

130.2 : 71 = 1.834

mol : coefficient of reactant : P₄ : Cl₂ :

= 0.3675/1 : 1.834/6

= 0.3675 : 0.306

Cl₂ as limiting reactant(smaller ratio)

Reacted mol of P₄ (as an excess reactant):

=1/6 x 1.834

= 0.306

Unreacted mol of P₄ :

= 0.3675 - 0.306

= 0.0615

Mass of P₄(left) :

= 0.0615 x 124

= 7.626 g