Elemental phosphorus reacts with chlorine gas according to the equation p4(s)+6cl2(g)→4pcl3(l) a reaction mixture initially contains 45.77 g p4 and 130.0 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.

The balanced equation for the above reaction is;
P₄ + 6Cl₂ ---> 4PCl₃
Stoichiometry of P₄ to 6Cl₂ is 1:6
every 1 mol of P₄ reacts with 6 mol of Cl₂.
First we need to calculate the number of moles reacted of both P₄ and Cl₂
Number of P₄ moles  - 45.77 g/ 124 g/mol = 0.369 mol
Number of Cl₂ moles - 130.0 g / 70.9 g/mol = 1.83 mol
limiting reactant is the reagent which is fully consumed up and amount of product formed depends on amount of limiting reactant present.
Excess reactant is present in excess and only a fraction of the amount is used up in the reaction. 
if P₄ is the limiting reactant,
if 1 mol  of P₄ reacts with 6 mol of Cl₂
then 0.369 mol reacts with - 6 x 0.369 = 2.21 mol
However only 1.83 mol of Cl₂ is remaining, that means P₄ is in excess.
1 mol of Cl₂ reacts with - 1/6 mol of P₄
then 1.83 mol of Cl₂ reacts with - 1/6 x 1.83 = 0.305 mol
0.369 mol of P₄ provided but only 0.305 mol reacted
Moles of P₄ in excess - 0.369 - 0.305 = 0.064 mol
mass of P₄ in excess - 0.064 mol x 124 g/mol = 7.94 g