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I'm not an expert teacher or lecturer of chemistry. I was only a student from SMA NEGERI 15 SURABAYA who had been one of the Bronze Medalist Participants of Olimpiade Sains Nasional X (2011) of Chemistry In Manado, North Sulawesi, 11 - 16 September 2011 and graduated in 2012. Now, I'm studying at Universitas Airlangga in Surabaya, Indonesia. I do love chemistry and I would like to help them who had difficulties in studying chemistry. That's why, please understand me if you found some misconcepts in my entries. Suggestions are always necessary in order to develop this blog. And I'm sorry because my English isn't so well.

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Saturday, December 19, 2009

Determining The Excess Reagent and The Limiting Reagent In a Chemical Equation.

In Senior High School especially at X grade, we will learn how to determine the excess and limiting reagent in a chemical formula and I’ve seen many friends of mine that confused because they can’t determine them.

So,

I wanna help you to determine the excess and limiting reagent, and I hope I tell you perfectly. Because I’m not a teacher, I’m a student like you, so maybe I wrong in this article.

Okay, here we go.

I’ll make an example of chemical reaction:

CO2(g) + H2O(l) --> O2(g) + C6H12O6(s)


Nah, we had a chemical reaction, and now, I’ll make examples of question:


1. Balance the equation!

2. Determine the volume of O2 and C6H12O6 if the volume of CO2 Is 200 liters and the volume of H2O is 250 liters at the same Temperature and Pressure! Don’t forget to determine the excess and the limiting reagent from that question too!

3. Determine the mass of O2 and C6H12O6 if the mass of CO2 is 30 g and the mass of H2O is 54 g at the same Temperature and Pressure! Don’t forget to determine the excess and the limiting reagent from that question too!

4. Determine the mole of O2 and C6H12O6 if the mole of H2O is 1.2 moles and the mole of CO2 is 3.6 moles! Don’t forget to determine the excess and the limiting reagent from that question too!

5. Determine the volume (based from question 2) after reaction!

6. Determine the mass (based from question 3) after reaction!

7. Expand that chemical reaction in a paragraph!


Now, let’s answer those questions!

Balance the equation…


CO2 + H2O --> O2 + C6H12O6


The First Step:

CO2 + H2O --> O2 + C6H12O6

The total amount of C in the left side is 1 and the total amount of C in the right side is 6 so?

Time the C in the left side with 6 (As the coefficient)!

Then our chemical reaction will change into:

6CO2 + H2O --> O2 + C6H12O6


How about the total amount of O?

Aaaaah, don’t care the O, the total amount of O in the left and the right side will same automatically when we finish to balance this equation.


The Second Step:

6CO2 + H2O --> O2 + C6H12O6

The total amount of H in the left side is 2 while the total amount of H in the right side is 12 so?

Time the H in the left side with 6 (Same like how we time the total amount of C)!

Tadaaaa!

6CO2 + 6H2O --> O2 + C6H12O6


Hmm… Now we already balance the C and H, now what we do next?

Yes you were right!


Let’s check our chemical reaction:

Left Side: C = 6, H = (2 x 6 =)12, and O = ((6 x 2) + 6 =) 18.

Right Side: C = 6, H = 12, and O = (2 + 6 =) 8.


Oooh no… the total amount of O aren’t same… what should we do now???

Hmmm… I know you know... :)

Time the O in the right side with 6!

So, let see our lovely chemical reaction!

6CO2 + 6H2O --> 6O2 + C6H12O6

Are we finish yet?

Of course! :D


Now let’s answer the question 2!

Determine the volume of O2 and C6H12O6 if the volume of CO2 Is 200 liters and the volume of H2O is 250 liters at the same Temperature and Pressure. Don’t forget to determine the excess and the limiting reagent from that question too…


6CO2 + 6H2O --> 6O2 + C6H12O6

200 liters 250 liters

Based on the Gay Lussac’s Law, the coefficient of a molecule in the same temperature and pressure is same with the mole of that molecule.


So:

CO2 : H2O : O2 : C6H12O6

6 : 6 : 6 : 1

Now we have a ratio, so what we do next?

Exactly!


CO2 : H2O : O2 : C6H12O6

6 moles : 6 moles : 6 moles : 1 mol

200 liters : 250 liters : ? liter(s) : ? liter(s)

-->Determine the limiting reagent!


CO2 :

200/6 = 33.3

H2O
250/6 = 41.6

Hmm… the value of CO2 is less than H2O so, can you conclude this?

Yes!

CO2 is the limiting reagent!


It’s easy right?

But what we do next?

Let’s see…


CO2:

33.3 x 6 moles = 200

200 liters – 200 = 0


H2O:

33.3 x 6 moles = 200

250 liters – 200 = 50 liters

--> The excess reagent is H2O with the excess is: 50 liters


Now let’s find the volume of O2 and C6H12O6 !

O2:

6/6 x 200 = 200 liters

C6H12O6

1/6 x 200 = 33.3 liters


Conclusion:

We use the ratio from the limiting reagent to determine the volume of the other molecules.


Now Let’s answer the question 3!

Determine the mass of O2 and C6H12O6 if the mass of CO2 is 30 g and the mass of H2O is 54 g at the same Temperature and Pressure! Don’t forget to determine the excess and the limiting reagent from that question too…

6CO2 + 6H2O --> 6O2 + C6H12O6

30 g 54 g


Remember:

Based on Lavoisier’s Law, the mass before reaction and the mass after reaction are same.

=

CO2 + H2O = O2 + C6H12O6

6 + 6 = 6 + 6

30 : 54 : ? g : ? g

PS: Numbers on the left side are based from the coefficient and the numbers on the left side is got from 6+6 = 12, 12:2 = 6

CO2:

30/6 = 5 --> The limiting reagent

H2O:

54/6 = 9 --> The excess reagent


=

CO2 : H2O : O2 : C6H12O6

6 : 6 : 6 : 6

30 : 54 : ? g : ? g

(5x6) : (5x6) : (5x6) : (5x6)


=

CO2 : H2O : O2 : C6H12O6

30 : 54 : ? g : ? g

30 : 30 : 30 : 30

----------------------------------------------- ­­___

0 : 24g : 30g : 30g


Conclusion:

1. The excess (From H2O) is 24 grams.

2. The mass of O2 is 30 grams.

3. The mass of C6H12O6 is 30 grams.


Now let’s answer the question 4!

Determine the mole of O2 and C6H12O6 if the mole of O2 is 1.2 moles and the mole of CO2 is 3.6 moles! Don’t forget to determine the excess and the limiting reagent from that question too…


Actually, the way to determine the mole is same like the way to determine the volume but don’t worry, I still

show you how to find the mole.. :)


6CO2 + 6H2O --> 6O2 + C6H12O6

3.6 moles 1.2 moles

=

CO2 :

3.6/6 = 0.6 --> The excess reagent

H2O
1.2/6 = 0.2 --> The limiting reagent


It’s easy right? :)

CO2:

0.2 x 6 = 1.2

3.6 moles – 1.2 = 2.4 moles

H2O:

0.2 x 6 = 1.2

1.2 moles – 1.2 = 0

--> The excess reagent is CO2 with the excess is: 2.4 moles


Now let’s find the mole of O2 and C6H12O6 !

O2:

6/6 x 1.2 = 1.2 moles

C6H12O6

1/6 x 1.2 = 0.2 moles


Conclusion:

1. The excess (CO2) is 2.4 moles

2. The mole of O2 is 1.2 moles

3. The mole of C6H12O6 is 0.2 moles


Now Let’s answer the question 5!

Determine the volume (based from question 2) after reaction….


The data from the answer of question 2:

1. CO2 is the limiting reagent which excess is: 50 liters.

2. O2 = 200 liters

3. C6H12O6 = 33.3 liters


Okay, we have the data, so…. What will happen???

The volume after reaction =

The volume of O2 + The volume of C6H12O6 + Excess =

200 liters + 33.3 liters + 50 liters = 283.3 liters.


WAIT A MINUTE!


The volume before reaction = the volume after reaction ?

450 liters = 283.3 liters ????


Hmm.. let’s back to our chemical reaction:

6CO2(g) + 6H2O(l) --> 6O2(g) + C6H12O6(s)

Look at the O2

What is the form of O2?

Gas? Okay you right!

And what is the relation between them?

Okay, this is the answer:

When the O2 formed, it will make the volume decrease, why?

Like when we cook water, is the gas still in its place? No, isn’t it?

And it same with this reaction. :)


Now for the question 6…

Determine the mass (based from question 3) after reaction…


Let see the data from the answer of question 3:

1. The excess (From H2O) is 24 grams.

2. The mass of O2 is 30 grams.

3. The mass of C6H12O6 is 30 grams.


The mass after reaction:

The mass of O2 + The mass of C6H12O6 + Excess =

30 grams + 30 grams + 24 grams = 84 grams.


Is the mass before and after reaction same?

Let’s see!

The mass before reaction = the mass after reaction

54 grams = 54 grams


See? It’s too easy right? :D


Okay, now let’s answer the last question! :)


Expand that chemical reaction in a paragraph…

6CO2 + 6H2O --> 6O2 + C6H12O6

=

6 molecules of Carbon dioxide (CO2) react with 6 molecules of Water (H2O) to form 6 molecules of Oxygen (O2) and a molecule of Carbohydrate (C6H12O6).


OKAY WE’VE FINISHED ALL OF THOSE QUESTIONS !!! :)


Now, you can try this question in your home for your training okay?

(I can’t believe, I made many of chemistry question by myself! Magic! :D)


1. Pb(s) + PbO2(aq) + 2H2SO4(aq) --> 2PbSO4(aq) + 2H2O(l)

(Ar Pb = 207, Ar O = 16, Ar H = 1, Ar S = 32)

a. The volume of PbO2 is 331 liters, what is the volume of the others at the same temperature and pressure?

b. The volume of Pb is 750 liters and the volume of 2H2SOis 500 liters, what is the volume of 2H2O?

c. If the Temperature is 32o C, the Pressure is 3 atm, and the mass of 2PbSO4 is 443 grams, determine it volume (Remember: PV=nRT, and n=m/Mr) !

d. If 2 liter of gas X equal to 0.75 mass (in grams) of 1.5 liter 2H2O (Mr 2H2O = 36) at the same temperature and pressure, determine the Ar of X! (Remember, At the same temperature and pressure: P1xV1/T1 = P2xV2/T2)

e. Expand that equation in a sentence!


2. Ca(s)+ 2H2O(l) --> Ca(OH)2(aq) + H2(g)

(Ar Ca = 40, Ar H = 1, Ar O = 16)

a. The mole of Ca is 3.4 moles, what is the mole of the others?

b. If the mole of Ca is 4 moles and the mole of 2H2O is 6 moles, what is the mole of Ca(OH)2 ?

c. Expand that equation in a sentence!


3. Fe(s) + O2(g) --> Fe2O3(g)

(Ar Fe = 56, Ar O = 16)

a. The mass of Fe is 75 kg, how about the others?

b. If the mass of Fe is 64 kg and the mass of O2 is 72 kg, determine the mass of Fe23!

c. Determine the mass after reaction!

d. Expand that equation in a sentence!



The Answer Key:


1 a. Pb: 331 liters, H2SO4: 662 liters, H2O: 662 liters.

b. 500 liters.

c. 6.09 liters.

d. 20.76 grams.

e. 1 molecule of Lead (solid) react with 1 molecule of Lead(II) oxide (aqueous) and with

2 molecules of Sulfuric acid and then form 2 molecules of Lead(II) sulfate (aqueous)

And 2 molecules of water.


2 a. 2H2O: 6.8 moles, Ca(OH)2: 3.4 moles, H2: 3.4 moles.

b. 3 moles.

c. 1 molecule of Calcium (solid) react with 2 molecules of water and then form 1

molecule of Calcium hydroxide (aqueous) and 1 molecule of Hydrogen (gas).


3 a. O2: 8.016 kg, Fe2O3: 53.44 kg.

b. 112 kg.

c. 136 kg.

d. 4 atoms of Fe (solid) reacts with 3 molecules of Oxygen (gas) to form 2 molecules of

Iron(III) oxide.


GOOD LUCK! :)

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