Answer letter D, E and F only. ll. Direction: Carefully read the following problemsDo the following :A. Balance the chemical equation B. Identify the limiting reactant and the excess reactant if applicable;C. Compute for the theoretical yield;D. Determine the percentage yield of the reaction;E. Calculate the percent error andF. Compute for the excess amount of the excess reactant if applicable; 3. Fifty grams of Chlorine reacts completely with Phosphorus obtaining ninety grams of theproduct.P +C12,PCI. Step 2Recall the reaction,P + ClPCI3----->mass of Cl2 = 50g,molar mass of Cl2=70.90 g mol1mass of PCI3 = 90 gMolar mass of PC|3= 137.33 g mol--Step 3A) Reaction,P + Cl2PCI3----->Balanced reaction,2P + 3C12-----> 2PCI3B) In this question, only mass of Cl2 is given so Cl2 act as limiting reagent because it is completelyconsumed and P act as excess reagent.Step 4C)From Balanced reaction,2P + 3C12-> 2PCI3----->3x70.90 g of Cl2 produced 2x137.33 g of PCI3=> 1 g of Cl2 produced = 1.29 g of PCI3so,50 g of Cl2 produced = 50x1.29 g of PCI3d - CAC

Given that 50 g of chlorine reacts completely with phosphors and 90 g of PCl3 is obtained according…

Answered: D. Determine the percentage yield of…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Answer letter D, E and F only.

ll. Direction: Carefully read the following problems
Do the following :
A. Balance the chemical equation
B. Identify the limiting reactant and the excess reactant if applicable;
C. Compute for the theoretical yield;
D. Determine the percentage yield of the reaction;
E. Calculate the percent error and
F. Compute for the excess amount of the excess reactant if applicable;

3. Fifty grams of Chlorine reacts completely with Phosphorus obtaining ninety grams of the
product.
P +C12,
PCI.

Step 2
Recall the reaction,
P + Cl
PCI3
----->
mass of Cl2 = 50g,
molar mass of Cl2=70.90 g mol1
mass of PCI3 = 90 g
Molar mass of PC|3= 137.33 g mol--
Step 3
A) Reaction,
P + Cl2
PCI3
----->
Balanced reaction,
2P + 3C12
-----> 2PCI3
B) In this question, only mass of Cl2 is given so Cl2 act as limiting reagent because it is completely
consumed and P act as excess reagent.
Step 4
C)From Balanced reaction,
2P + 3C12
-> 2PCI3
----->
3x70.90 g of Cl2 produced 2x137.33 g of PCI3
=> 1 g of Cl2 produced = 1.29 g of PCI3
so,
50 g of Cl2 produced = 50x1.29 g of PCI3
d - CAC

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