Question

Consider the reaction: P_​4 ​+ 6Cl_​2 ​-> 4PCl₃. a. How many grams of Cl_​2 are needed to react with 20.00 g. of P₄​? __________ b. You have 15.00 g. of P​₄ and 22.00 g. of Cl₂​, identify the limiting reactant and calculate the grams of PCl₃​ that can be produced as well as the grams of excess reactant remaining. LR _________ grams PCl_{​3 ___________} grams excess reactant ________ c. If the actual amount of PCl₃ ​recovered is 16.25 g., what is the percent yield? ________ d. Given 28.00 g. of P₄ and 106.30 g. of Cl₂​, identify th limiting reactant and calculate how many grams of excess reactant will remain after the reaction. LR _________ grams excess reactant _________

Answer 1

P​4 ​+ 6Cl​2 ​-> 4PCl3

atomic weights : P = 30.97 g /mol ==> 4P = 4*30.97

Cl = 35.45 g/ mol ==> Cl2 = 2 * 35.45

a)

Given, 20.00 g of P4

therefore, no. of moles of P4 = weight of P4 / molar mass of P4 = 20.00 / 4* 30.97 = 0.161 moles

from the stoichiometric equation, 1 mole of P4 requires -------> 6 moles of Cl2

==> 0.161 moles of P4 requires --------> 6 * 0.161 = 0.96 moles of Cl2

Therefore, mass of Cl2 = no. of moles * molar mass of Cl2 = 0.96 * (2 * 35.45) = 68.06 g of Cl2

Mass of Cl2 needed = 68.06 g of Cl2.

b)

mass of P4 = 15.00g and mass of Cl2 = 22.00 g

no.of moles of P4 = weight / molar mass = 15.00 / 4* 30.97 = 0.12 moles of P4

no. of moles of Cl2 = weight / molar mass = 22.00 / (2*35.45 )= 0.31 moles of Cl2

from the stoichiometric equation,  1 mole of P4 requires -------> 6 moles of Cl2

==> 0.12 moles of P4 requires --------> 6 * 0.12 = 0.72 moles of Cl2

but we have only 0.31 moles of Cl2 that means Cl2 is less in amount than theoretically required.

Therefore, Cl2 is the limiting reactant. and Cl2 required to complete the reaction is = 0.72- 0.31 = 0.41 moles

0.41 moles = 0.41 * (35.45*2) = 29.06 g of Cl2 is required.

and P4 excess in amount.

from the stoichiometric equation, 6 moles of Cl2 requires ----> 1 mole of P4

0.31 moles of Cl2 requires ----> (1/6) * 0.31 = 0.05 moles of P4

0.05 moles of P4 = 0.05 * 4* 30.97 = 6.4 g of P4 reacted in the reaction.

Therefore, excess amount of P4 = 15.00- 6.4 = 8.60 g in excess of P4

Now, PCl3

1 mole of P4 produces ----> 4 moles of PCl3

  0.05 moles of P4 produces -----> 4*0.05 = 0.20 moles PCl3 produced.

0.20 moles of PCl3 = 0.20* ( 30.97 + 3* 35.45 ) = 27.46 g of PCl3

Therefore, Moles of PCl3 produced = 27.46 g of PCl3

c)

Actual amount of PCl3 recovered = 16.25 g

Theoretical amount of PCl3 = 27.46 g

Yield = (actual amount/ theoretical amount ) *100 = (16.25/27.46) *100 = 59.18%

Therefore, Yield = 58% of PCl3

d)

mass of P4 = 28.00g and mass of Cl2 = 106.30 g

no.of moles of P4 = weight / molar mass = 28.00 / 4* 30.97 = 0.23 moles of P4

no. of moles of Cl2 = weight / molar mass = 106.30 / (2*35.45 )= 1.50 moles of Cl2

from the stoichiometric equation,  1 mole of P4 requires -------> 6 moles of Cl2

==> 0.23 moles of P4 requires --------> 6 * 0.23 = 1.38 moles of Cl2

but we have 1.50 moles of Cl2 that means Cl2 is more in amount than theoretically required.

Therefore, P4 is the limiting reactant. and Cl2 excess in the reaction is = 1.50- 1.38 = 0.12 moles

0.12 moles = 0.12 * (35.45*2) = 8.51 g of Cl2 is in excess