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.)

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

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