Answer:
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
