Question

For the following reaction, 23.3 grams of phosphorus (P) are allowed to react with 84.8 grams of chlorine gas. phosphorus (P) (s) + chlorine (g) phosphorus trichloride () What is the maximum amount of phosphorus trichloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer

Answer 1

balanced reaction is

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

given
mass of P4 = 23.3g

molar mass of P4 = 124.895 g/mol

moles of P4 = given mass/molar mass = 23.3g/124.895 g/mol = 0.186 mol

mass of Cl2 =84.8 g

molar mass of Cl2 = 70.9 g/mol

moles of Cl2 = 84.8g/70.9 g/mole = 1.194 mole

According to balanced reaction 1 mole P4 needs 6 moles of Cl2

here 0.186 mols P4 require 6*0.186mol Cl2 = 1.116

but we have 1.194 mol of Cl2 hence Cl2 will be excess reagent and P4 is the limiting reagent

since limiting reactant is P4,we need to calculate mols of P4 to clculate mol of product

From balanced equation 1mols P4 produce 4 mols PCl3
So 0.186 mols P4 yields 4 * 0.186 = 0.751 mols PCl3

now convert mols to mass

mass of PCl3 = mols of PCl3*molar mass =0.751*137.33 g/mol
= 103.13 g

moles of Cl2 left =

0.187 moles of P4 react with 1.116 moles of Cl2

moles of Cl2 left = 1.194 -1.116 = 0.078 moles

mass of Cl2 unreacted= 0.078 mols*71g/mol = 5.538g

PLEASE LEAVE THE POSITIVE FEEDBACK IF YOU LIKE THE SOLUTION