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:

CO_{2}(g)^{ }+ H_{2}O(l) --> O_{2}(g) + C_{6}H_{12}O_{6}(s)

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

1. Balance the equation!

2. Determine the volume of O_{2} and C_{6}H_{12}O_{6} if the volume of CO_{2} Is 200 liters and the volume of H_{2}O 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 O_{2 }and C_{6}H_{12}O_{6 }if the mass of CO_{2} is 30 g and the mass of H_{2}O 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 O_{2} and C_{6}H_{12}O_{6} if the mole of H_{2}O is 1.2 moles and the mole of CO_{2} 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…*

CO_{2}^{ }+ H_{2}O --> O_{2} + C_{6}H_{12}O_{6}

*
*

*The First Step:*

** C**O

_{2}

^{ }+ H

_{2}O --> O

_{2}+

**H**

*C*_{6}_{12}O

_{6}

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:

** 6C**O

_{2}

^{ }+ H

_{2}O --> O

_{2}+

**H**

*C*_{6}_{12}O

_{6}

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:*

6CO_{2}^{ }+ ** H_{2}**O --> O

_{2}+ C

_{6}

**O**

*H*_{12}_{6}

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!

6CO_{2}^{ }+ ** 6H_{2}**O --> O

_{2}+ C

_{6}

**O**

*H*_{12}_{6}

_{ }

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!

** 6**C

*O*_{2}^{ }+

**H**

*6*_{2}

**-->**

*O***+ C**

*6O*_{2}_{6}H

_{12}

*O*_{6}_{ }

Are we finish yet?

Of course! :D

Now let’s answer the question 2!

*Determine the volume of O _{2} and C_{6}H_{12}O_{6} if the volume of CO_{2} Is 200 liters and the volume of H_{2}O is 250 liters at the same Temperature and Pressure. Don’t forget to determine the excess and the limiting reagent from that question too…*

6CO_{2}^{ }+ 6H_{2}O --> 6O_{2} + C_{6}H_{12}O_{6}

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:

CO_{2}^{ }: H_{2}O : O_{2} : C_{6}H_{12}O_{6}

6 : 6 : 6 : 1

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

Exactly!

CO_{2}^{ }: H_{2}O : O_{2} : C_{6}H_{12}O_{6}

6 moles : 6 moles : 6 moles : 1 mol

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

-->Determine the limiting reagent!

CO_{2} :

H_{2}O

Hmm… the value of CO_{2} is less than H_{2}O so, can you conclude this?

Yes!

**CO_{2} is the limiting reagent**!

It’s easy right?

But what we do next?

Let’s see…

CO_{2}:

33.3 x 6 moles = 200

200 liters – 200 = 0

H_{2}O:

33.3 x 6 moles = 200

250 liters – 200 = 50 liters

__-->____ The excess reagent is H _{2}O with the excess is: 50 liters__

Now let’s find the volume of O_{2} and C_{6}H_{12}O_{6} !

O_{2}:

C_{6}H_{12}O_{6}

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 O _{2 }and C_{6}H_{12}O_{6 }if the mass of CO_{2} is 30 g and the mass of H_{2}O is 54 g at the same Temperature and Pressure! Don’t forget to determine the excess and the limiting reagent from that question too…*

6CO_{2}^{ }+ 6H_{2}O --> 6O_{2} + C_{6}H_{12}O_{6}

30 g 54 g

Remember:

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

=

CO_{2}^{ }+ H_{2}O = O_{2} + C_{6}H_{12}O_{6}

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

CO_{2}:

H_{2}O:

=

CO_{2}^{ }: H_{2}O : O_{2} : C_{6}H_{12}O_{6}

6 : 6 : 6 : 6

30 : 54 : ? g : ? g

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

=

CO_{2}^{ }: H_{2}O : O_{2} : C_{6}H_{12}O_{6}

30 : 54 : ? g : ? g

30 : 30 : 30 : 30

----------------------------------------------- _{}^{___}

^{ } 0 : 24g : 30g : 30g

Conclusion:

1. The excess (From H_{2}O) is 24 grams.

2. The mass of O_{2} is 30 grams.

3. The mass of C_{6}H_{12}O_{6 }is 30 grams.

Now let’s answer the question 4!

*Determine the mole of O _{2} and C_{6}H_{12}O_{6} if the mole of O_{2} is 1.2 moles and the mole of CO_{2} 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.. :)

6CO_{2}^{ }+ 6H_{2}O --> 6O_{2} + C_{6}H_{12}O_{6}

3.6 moles 1.2 moles

=

CO_{2} :

H_{2}O

It’s easy right? :)

CO_{2}:

0.2 x 6 = 1.2

3.6 moles – 1.2 = 2.4 moles

H_{2}O:

0.2 x 6 = 1.2

1.2 moles – 1.2 = 0

__-->____ The excess reagent is CO _{2} with the excess is: 2.4 moles__

Now let’s find the mole of O_{2} and C_{6}H_{12}O_{6} !

O_{2}:

C_{6}H_{12}O_{6}

Conclusion:

1. The excess (CO_{2}) is 2.4 moles

2. The mole of O_{2} is 1.2 moles

3. The mole of C_{6}H_{12}O_{6 }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. CO_{2} is the limiting reagent which excess is: 50 liters.

2. O_{2 }=_{ }200 liters

3. C_{6}H_{12}O_{6 }= 33.3 liters

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

The volume after reaction =

The volume of O_{2} + The volume of C_{6}H_{12}O_{6 }+ 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:

6CO_{2}(g)^{ }+ 6H_{2}O(l) --> 6O_{2}(g) + C_{6}H_{12}O_{6}(s)

Look at the O_{2}…

What is the form of O_{2}?

Gas? Okay you right!

And what is the relation between them?

Okay, this is the answer:

When the O_{2} 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 H_{2}O) is 24 grams.

2. The mass of O_{2} is 30 grams.

3. The mass of C_{6}H_{12}O_{6 }is 30 grams.

The mass after reaction:

The mass of O_{2} + The mass of C_{6}H_{12}O_{6 }+ 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…*

6CO_{2}^{ }+ 6H_{2}O --> 6O_{2} + C_{6}H_{12}O_{6}

_{ }

=

*6 molecules of Carbon dioxide (CO _{2}) _{ }react with 6 molecules of Water (H_{2}O) to form 6 molecules of Oxygen (O_{2}) and a molecule of Carbohydrate (C_{6}H_{12}O_{6}).*

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) + PbO_{2}(aq) + 2H_{2}SO_{4}(aq) --> 2PbSO_{4}(aq) + 2H_{2}O(l)

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

a. The volume of PbO_{2 }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 2H_{2}SO_{4 }is 500 liters, what is the volume of 2H_{2}O?

c. If the Temperature is 32^{o} C, the Pressure is 3 atm, and the mass of 2PbSO_{4} is 443 grams, determine it volume (Remember: _{PV=nRT, }and

d. If 2 liter of gas X equal to 0.75 mass (in grams) of 1.5 liter 2H_{2}O (Mr 2H_{2}O = 36) at the same temperature and pressure, determine the Ar of X! (Remember, At the same temperature and pressure:

e. Expand that equation in a sentence!

2. Ca(s)+ 2H_{2}O(l) --> Ca(OH)_{2}(aq) + H_{2}(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 2H_{2}O is 6 moles, what is the mole of Ca(OH)_{2} ?

c. Expand that equation in a sentence!

3. Fe(s) + O_{2}(g) --> Fe_{2}O_{3}(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 O_{2} is 72 kg, determine the mass of Fe_{2}O_{3}!

c. Determine the mass after reaction!

d. Expand that equation in a sentence!

*
*

*
*

*The Answer Key:*

1 a. Pb: 331 liters, H_{2}SO_{4}: 662 liters, H_{2}O_{: }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. 2H_{2}O: 6.8 moles, Ca(OH)_{2}: 3.4 moles, H_{2}: 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. O_{2}: 8.016 kg, Fe_{2}O_{3}: 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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