Elemental phosphorus reacts with chlorine gas according to the equation: p4(s)+6cl2(g)❝4pcl3(l) a reaction mixture initially contains 45.55g p4 and 131.9g cl2.

once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

So after reaction completion. 7.11 g of P₄ will be left over.  

Explanation:

So for number of moles of P₄ =    Given mass/ molar mass = 45.55/ 123.88

                                                                                      = 0.367 moles  

Number of moles of Cl₂ = Given mass/ molar mass = 131.9/70 = 1.884 moles  

We will divide the values by dividing excess of reactants with their coefficient for finding excess of reactants.  

P₄ = 0.367/ coefficient 1 = 0.367

Cl₂ =   1.884/ 6= 0.314

Phosphorus is thus the excess reactant.

We need to use the mole value of the limiting reactant in order to find the mass of P₄ used. Using dimensional analysis,

1.884 mol Cl₂  ( 1mole P₄/6 mol Cl₂  )  ( 123.88 g P₄/ 1 mol P₄)   = 38.44 g P₄

This is the amount which is used up. So by subtracting this amount from the Initial amount  

45.55 – 38.44 =  7.11 g of P₄

So after reaction completion. 7.11 g of P₄ will be left over.