Consider the reaction: P4 + 6Cl2 = 4PCl3.
a. How many grams of Cl2 are needed to react with 20.00 g of P4? ______
b. You have 15.00 g. of P4 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 PCl3 _ 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 the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______ grams excess reactant
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b. You have 15.00 g. of P4 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 PCl3 _________ grams excess reactant ___________

P4 + 6Cl2 = 4PCl3

124g 426 g 4(31 + 3(35.5)) = 550g

15g 22g

I will use P4 to find the limiting reactant

x = (15 x 426) / 124 = 51.5 The limiting reactant is Chlorine

because we need 51.5 g and we only have 22g

Excess reactant

x = (22 x 124) / 426 = 6.4 g of P4

Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

426 g of Cl2 ---------------- 550 g of PCl3

22g of Cl2 ------------- - x

x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

% yield = (16.25 - 28.4) / 28.4 x 100

% yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

124g of P4 ------------- 426 g 6Cl2

28g --------------- x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3 - 96.2 = 10.1 g of Cl2 in excess