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. 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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The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g

                                P4      +      6Cl2      =      4PCl3

                          4(31) 12(35.5)

                         20        x

                    x = 20(12x35.5) / 4(31)

                   x = 8520 / 124

                   x = 68.7 g

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