P4(l) + 6Cl2(g) --------------> 4PCl3(l) no of moles of P4 = W/G.M.Wt = 100/124

Question

P4(l) + 6Cl2(g) --------------> 4PCl3(l)

no of moles of P4   = W/G.M.Wt

                              = 100/124   = 0.81 moles

no of moles of Cl2    = W/G.M.Wt

                                 = 320/71   = 4.5moles

1 moles of P4 react with 6 moles of Cl2

0.81 moles of P4 react with = 6*0.81/1   = 4.86 moles of Cl2

Cl2 is limiting reactant

6 moles of Cl2 react with P4 to gives 4 moles of PCl3

4.5 moles of Cl2 react with P4 to gives = 4*4.5/6 = 3 moles of PCl3

mass of PCl3 = no of moles * gram molar mass

                       = 3*137.5   = 412.5g of PCl3

Explanation / Answer

) Stoichiometry and Limiting Reagents (5 marks total): Consider the following balanced reaction P. (I) + 6C12(g) 4PC13(1) where 100 g of P4 are mixed with 320 g of Cl2 Use calculations to identify the limiting reagent (3 marks) Calculate the mass of PCl3 produced in grams (2 marks).

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