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;

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Chapter3: Stoichiometry
Section: Chapter Questions
Problem 13ALQ: Chlorine exists mainly as two isotopes, 37Cl and 33Cl. Which is more abundant? How do you know?
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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.
Transcribed Image Text: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
Transcribed Image Text: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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