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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?

Answer :

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Answer:

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.  

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