Stoikiometri (6) | Pereaksi Pembatas | Kimia Kelas 10

Kimatika
21 Apr 202112:40

Summary

TLDRThis educational video script focuses on stoichiometry for 10th-grade chemistry students. It explains the concept of limiting reactants in chemical reactions, where the reactant that is completely consumed dictates the reaction's outcome. The script guides through determining the limiting reactant by comparing initial moles to their respective coefficients. It then uses examples, such as the combustion of methane and the reaction of iron with sulfuric acid, to illustrate calculations for product mass and remaining reactants. The video concludes with a problem-solving approach to find the volume of hydrogen gas produced under standard conditions.

Takeaways

  • πŸ” The video discusses stoichiometry, focusing on the concept of limiting reactants in chemical reactions.
  • πŸ“š A limiting reactant is a substance that gets completely consumed during a reaction, determining the amount of product formed.
  • πŸ§ͺ The script explains two scenarios: one where all reactants are completely consumed and another where one reactant is completely consumed while the other is left over.
  • πŸ“‰ To determine the limiting reactant, compare the initial moles of reactants to their stoichiometric coefficients; the one with the smallest ratio is the limiting reactant.
  • πŸ”’ The script provides a step-by-step calculation for a chemical reaction involving the combustion of methane (CH4) and oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
  • βš–οΈ The mass of reactants (CH4 and O2) is converted to moles using their molar masses, which is crucial for determining the limiting reactant.
  • πŸ”„ The script demonstrates how to balance chemical equations, which is necessary for accurately determining the stoichiometry of a reaction.
  • πŸ“Š The video uses an example to show how to calculate the mass of CO2 produced from the combustion of CH4 and the remaining mass of O2.
  • πŸ”— The concept of molar ratios is applied to determine the amount of product formed and the amount of reactant left over after the reaction.
  • 🌐 The video concludes with another example involving the reaction of iron (Fe) with sulfuric acid (H2SO4) to produce iron sulfate and hydrogen gas, illustrating the application of stoichiometry in different contexts.

Q & A

  • What is stoichiometry in chemistry?

    -Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions.

  • What are the two possibilities when substances react?

    -There are two possibilities when substances react: 1) All reactants react completely, indicated by the initial moles being equal to the stoichiometric coefficients, and 2) One reactant is completely consumed while the other is left over, indicated by the initial moles not being equal to the stoichiometric coefficients.

  • What is the term used for a reactant that limits the outcome of a reaction?

    -The reactant that limits the outcome of a reaction is called the limiting reactant or, in short, the 'pepeya'.

  • How do you determine which reactant is the limiting reactant?

    -To determine the limiting reactant, compare the moles of each reactant with their respective stoichiometric coefficients. The reactant with the smallest ratio of initial moles to its coefficient is the limiting reactant.

  • Why is it important to determine the limiting reactant in a reaction?

    -Determining the limiting reactant is important because it provides a reference for calculating the amount of other reactants or products in a chemical reaction.

  • What is the balanced chemical equation for the combustion of methane (CH4) with oxygen (O2)?

    -The balanced chemical equation is CH4 + 2O2 β†’ CO2 + 2H2O, where methane reacts with oxygen to produce carbon dioxide and water.

  • How do you calculate the molar mass of methane (CH4) and oxygen (O2)?

    -The molar mass of methane (CH4) is calculated as 12 (for carbon) + 4*1 (for hydrogen) = 16 g/mol. For oxygen (O2), it is 2*16 (for oxygen) = 32 g/mol.

  • What is the mass of CO2 produced from the combustion of 3.2 grams of methane (CH4)?

    -The mass of CO2 produced is 8.8 grams, calculated by using the stoichiometry of the reaction and the molar mass of CO2 (44 g/mol).

  • How do you find the remaining mass of oxygen (O2) after the reaction with methane (CH4)?

    -To find the remaining mass of oxygen, first calculate the moles of O2 that reacted using the stoichiometry of the reaction, then subtract this from the initial moles of O2 to find the remaining moles, and finally convert this to mass using the molar mass of O2.

  • What is the balanced chemical equation for the reaction of iron (Fe) with sulfuric acid (H2SO4) to produce iron sulfate and hydrogen gas?

    -The balanced chemical equation is Fe + H2SO4 β†’ FeSO4 + H2, where iron reacts with sulfuric acid to produce iron sulfate and hydrogen gas.

  • How do you calculate the volume of hydrogen gas (H2) produced at standard temperature and pressure (STP) from a given reaction?

    -To calculate the volume of hydrogen gas at STP, first determine the moles of H2 produced using the stoichiometry of the reaction, then use the molar volume of a gas at STP (22.4 L/mol) to find the volume.

Outlines

00:00

πŸ” Stoichiometry and Reaction Limiting Factors

This paragraph introduces the concept of stoichiometry in chemistry, specifically focusing on the limiting reactant in a chemical reaction. It explains that a reaction can either be complete or limited by one of the reactants. The limiting reactant is determined by comparing the initial moles of reactants to their stoichiometric coefficients. The paragraph provides a method to identify the limiting reactant by dividing the initial moles of each reactant by their respective coefficients, with the smallest value indicating the limiting reactant. The importance of determining the limiting reactant is highlighted as it serves as a reference for calculating the amount of other reactants or products formed. The paragraph concludes with an example of a chemical reaction involving the combustion of methane (CH4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), illustrating the steps to determine the limiting reactant and the subsequent calculation of the products formed.

05:02

πŸ”¬ Calculation of Reaction Products and Remaining Reactants

The second paragraph delves into the calculation of products and remaining reactants in a chemical reaction. It uses the example of methane combustion to explain how to determine the mass of CO2 produced and the remaining reactants. The process involves calculating the moles of reactants using their mass and molar mass, identifying the limiting reactant, and then using stoichiometric ratios to find the moles of products. The paragraph demonstrates the calculation of the mass of CO2 formed by multiplying the moles of the limiting reactant by the molar mass of CO2. It also explains how to find the remaining moles of O2 by subtracting the moles of O2 that reacted from the initial moles. The paragraph provides a step-by-step approach to solving stoichiometry problems, emphasizing the importance of understanding the limiting reactant in predicting the outcome of chemical reactions.

10:04

🌐 Determination of Hydrogen Gas Volume in Standard Conditions

The final paragraph discusses the calculation of the volume of hydrogen gas (H2) produced in a reaction under standard temperature and pressure (STP). It uses the reaction between iron (Fe) and sulfuric acid (H2SO4) to form iron sulfate and hydrogen gas as an example. The paragraph explains how to determine the volume of hydrogen gas by first calculating the moles of hydrogen gas using stoichiometric relationships and the moles of the limiting reactant. It highlights the process of identifying the limiting reactant by comparing the moles of reactants to their stoichiometric coefficients. The calculation of the moles of hydrogen gas is demonstrated by using the moles of the limiting reactant and the stoichiometric coefficient. The paragraph concludes with the calculation of the volume of hydrogen gas at STP using the molar volume of a gas at STP (22.4 L/mol). This section emphasizes the practical application of stoichiometry in determining the volume of gases produced in chemical reactions.

Mindmap

Keywords

πŸ’‘Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It is crucial for determining the amounts of substances involved in reactions based on their chemical formulas. In the video, stoichiometry is used to calculate the limiting reactant and the amounts of products formed in chemical reactions, such as the combustion of methane (CH4) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

πŸ’‘Limiting Reactant

The limiting reactant is the substance that is completely consumed during a chemical reaction and thus determines the amount of product that can be formed. It is identified by comparing the mole ratio of the reactants to the stoichiometric coefficients in the balanced chemical equation. In the script, methane (CH4) is identified as the limiting reactant in the combustion reaction with oxygen, which helps in calculating the mass of CO2 produced.

πŸ’‘Balanced Equation

A balanced chemical equation represents the reaction where the number of atoms of each element on the reactant side equals the number on the product side, adhering to the law of conservation of mass. The script mentions the process of balancing the equation for the reaction between methane and oxygen, ensuring that the number of atoms for each element is equal on both sides.

πŸ’‘Mole Ratio

The mole ratio in a chemical reaction is the relationship between the amounts of reactants and products as expressed in moles, which is derived from the balanced chemical equation. The video uses mole ratios to determine the limiting reactant and to calculate the amount of products formed, such as the ratio between methane and oxygen in the combustion reaction.

πŸ’‘Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is used to convert between mass and the amount of substance in moles. In the script, molar mass is used to convert the mass of methane (CH4) and oxygen (O2) into moles, which is essential for stoichiometric calculations.

πŸ’‘Combustion Reaction

A combustion reaction is a redox chemical reaction between a substance and an oxidant, usually oxygen, that produces heat and light. The video discusses the combustion of methane (CH4) with oxygen (O2) as an example of a combustion reaction, resulting in the formation of carbon dioxide (CO2) and water (H2O).

πŸ’‘Oxidation State

Oxidation state, or oxidation number, is a measure used in chemistry to describe the degree of oxidation of an atom in a chemical compound. It is used in balancing redox reactions. Although not explicitly mentioned in the script, the concept is implicit in the discussion of the combustion reaction where methane is oxidized, and oxygen is reduced.

πŸ’‘Empirical Formula

An empirical formula represents the simplest whole number ratio of atoms in a compound. The script does not directly mention empirical formulas, but the concept is related to the discussion of balancing chemical equations, where the empirical formula helps determine the stoichiometric coefficients.

πŸ’‘Mole Concept

A mole is a unit of measurement used in chemistry to express amounts of a chemical substance, defined as the number of atoms in 12 grams of carbon-12 isotope, which is approximately 6.022 Γ— 10^23 entities. The video uses the mole concept to calculate the amounts of reactants and products in chemical reactions, such as determining the moles of methane and oxygen involved in the combustion reaction.

πŸ’‘Standard Temperature and Pressure (STP)

STP is a standard set of conditions for temperature and pressure (0Β°C and 1 atm) at which certain physical properties, like the volume of a gas, are measured. The video refers to STP when calculating the volume of hydrogen gas produced in a reaction, using the molar volume of an ideal gas at STP (22.4 liters per mole).

πŸ’‘Redox Reaction

A redox (reduction-oxidation) reaction involves the transfer of electrons between chemical species, typically involving a change in oxidation states. The combustion reaction discussed in the video is an example of a redox reaction, where methane is oxidized, and oxygen is reduced to form water and carbon dioxide.

Highlights

Introduction to stoichiometry in chemistry for class 10 students.

Explanation of complete and limiting reactions in stoichiometry.

Definition of limiting reactant and its role in determining the outcome of a reaction.

Method to determine the limiting reactant by comparing moles of reactants to their coefficients.

Importance of identifying the limiting reactant for calculating the amount of other reactants or products.

Example problem involving the combustion of methane (CH4) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

Step-by-step approach to balance the chemical equation for the combustion of methane.

Calculation of moles of CH4 and O2 from given masses using molar mass.

Determination of the limiting reactant between CH4 and O2 using their moles and coefficients.

Calculation of the mass of CO2 produced using the moles of the limiting reactant.

Determination of the remaining reactant and its mass after the reaction.

Introduction to a second example problem involving the reaction of iron (Fe) with sulfuric acid (H2SO4).

Balancing the chemical equation for the reaction between iron and sulfuric acid.

Calculation of moles of Fe from given mass and moles of H2SO4 from given volume.

Determination of the limiting reactant in the reaction between Fe and H2SO4.

Calculation of the volume of hydrogen gas (H2) produced at standard temperature and pressure.

Final thoughts and closing remarks for the video on stoichiometry.

Transcripts

play00:00

Halo Assalamualaikum Halo adik-adik

play00:02

ketemu lagi dengan Kakak di channel

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kinematika di video kali ini kita akan

play00:07

belajar materi kimia kelas 10 yaitu

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tentang stoikiometri dimana pada bagian

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ini yang akan kita bahas adalah pereaksi

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pembatas bila zat-zat direaksikan akan

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ada dua kemungkinan yang pertama adalah

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semua pereaksi tepat habis bereaksi nah

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ini ditandai dengan perbandingan mol

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awal atau mol mula-mula sama dengan

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perbandingan koefisien lalu kemungkinan

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yang kedua adalah salah satu pereaksi

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habis dan reaksi yang lain bersisa yang

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ditandai dengan perbandingan mol awal

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atau mula-mula tidak sama dengan

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perbandingan koefisien nah kemudian

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pereaksi yang habis bereaksi inilah yang

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akan membatasi hasil reaksi sehingga

play00:58

reaksi yang habis break

play01:00

wuih disebut juga dengan pereaksi

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pembatas kita singkat dengan pepeya lalu

play01:07

Bagaimana cara menentukan reaksi yang

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habis bereaksi atau pereaksi pembatas

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caranya adalah kita bandingkan jumlah

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mol pereaksi dengan koefisien

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masing-masing nah yang nilainya paling

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kecil itulah yang termasuk pereaksi

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pembatas Nah untuk apa sebenarnya kita

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menentukan pereaksi pembatas reaksi

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pembatas harus kita tentukan dulu untuk

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menjadi acuan dalam menentukan jumlah

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zat yang bereaksi lainnya atau jumlah

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zat yang dihasilkan JAdi misal dalam

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reaksi kimia molt-pre aksinya semua

play01:44

diketahui maka untuk menentukan Mal

play01:47

hasil reaksi yang menjadi acuan adalah

play01:50

Mal pereaksi yang habis bereaksi atau

play01:53

molt-pre aksi pembatas

play01:55

Hai Nah untuk lebih jelasnya kita akan

play01:58

membahas contoh soal dari

play02:01

di sejumlah 3,2 gram gas ch4 dibakar

play02:05

dengan 16 G gas O2 menurut reaksi

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berikut ch4 ditambah O2 menghasilkan CO2

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+ H2O diketahui air ha = 1 Arc = 12 dan

play02:19

Aero = 16 Tentukan a pereaksi pembatas

play02:24

bemassa CO2 yang terbentuk dan C reaksi

play02:28

sisa dan berapa sisanya nah

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langkah-langkah penyelesaiannya hampir

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sama dengan perhitungan pada reaksi

play02:34

kimia ya Yang pertama adalah kita buat

play02:37

dulu persamaan reaksi yang sudah setara

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ch4 ditambah O2 menghasilkan CO2 DIY

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tambah h2o ini persamaan reaksinya belum

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setara jadi kita setarakan dulu atom c

play02:54

di sebelah kiri satu di sebelah kanan

play02:56

satu sudah sama Hadi sebelah kiri 4 di

play02:59

sebelah kanan dua berarti sebelah kan

play03:01

terjadi empat koefisiennya kita ganti2

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kemudian kita samakan jumlah atom o o

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ikuti yang di sebelah kanan ya ada dua

play03:11

tambah dua jadinya 4 berarti sebelah

play03:13

kiri juga harus 4 supaya 4 koefisien

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kita ganti2 berikutnya kita tulis yang

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diketahui yaitu 3,2 gram gas ch4 jadi

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ch4 nya 3,2

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Hai kemudian 16 G gas O2 prastyo 2-nya

play03:34

16

play03:38

hai lalu kita cari mol masing-masing

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ingat cara mencari mol jika massa

play03:43

diketahui adalah dibagi dengan MR jadi

play03:47

3,2 dibagi dengan mr-nya si ch4 yaitu 16

play03:55

a = 0,2 mol

play03:59

Hai Kemudian untuk O2 dibagi dengan

play04:02

mr-nya sio2 yaitu 32 = 0,5 mol

play04:09

Hai ada dua mal yang diketahui berarti

play04:13

kita harus pilih Mall mana yang akan

play04:15

menjadi acuan dalam menentukan Mal atau

play04:19

masanya sih CO2 nah mal yang menjadi

play04:23

acuan adalah mol pereaksi pembatas cara

play04:26

menentukan reaksi pembatas mol pereaksi

play04:29

masing-masing dibagi dengan koefisiennya

play04:32

yang nilainya lebih kecil itulah yang

play04:35

menjadi reaksi pembatas kita bagi ya

play04:40

Hai 0,2 dibagi

play04:47

Hai kemudian 0,5 dibagi koefisiennya CO2

play04:52

yaitu 0,25 0,2 lebih kecil dibanding

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0,25 berarti yang menjadi pereaksi

play05:02

pembatas adalah

play05:06

Hai jadi jawaban yang a-a-a-a dalah ch4

play05:13

Hai jelas ya sekarang kita lanjut ke b

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masa CO2 yang terbentuk Nah untuk

play05:19

menentukan masa CO2 berarti kita harus

play05:23

cari mallnya CO2 dulu menggunakan

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perbandingan koefisien yaitu mol CO2 =

play05:31

koefisien CO2 satu per kofesien yang

play05:37

maunya diketahui berarti kita gunakan

play05:39

mallnya si pereaksi pembatas ya yaitu

play05:42

satu dikali dengan mallnya pereaksi

play05:46

pembatas yaitu

play05:50

a = 0,2 mol

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Hai telah kita peroleh mallnya barulah

play05:57

kita bisa mencari massa dari CO2 ingat

play06:01

untuk mencari massa atau G rumusnya

play06:04

adalah mal di kali Mr mallnya CO2 adalah

play06:11

0,2 mr-nya adalah 44 0,2 dikali 44 = 8,8

play06:21

kram

play06:24

Hai jadi masa CO2 yang terbentuk adalah

play06:28

8,8

play06:31

hai hai

play06:33

Hai berikutnya untuk yang c yang ditanya

play06:36

adalah pereaksi sisa dan masanya

play06:38

maksudnya adalah masa sisanya caranya

play06:42

kita cari dulu Mal O2 yang habis

play06:45

bereaksi menggunakan perbandingan

play06:47

koefisien juga ya berarti no2 mallnya

play06:52

CO2 = koefisiennya O2 yaitu 2per coffee

play06:57

senyap reaksi pembatas yaitu satu dikali

play07:00

dengan mallnya pereaksi pembatas yaitu

play07:03

0,2 sehingga diperoleh 0,4 mol

play07:10

Hai malu2 mula-mula 0,5 Mol yang

play07:13

bereaksi 0,4 mol berarti Sisanya adalah

play07:18

yang si Saudi wilayah sisa O2 adalah 0,5

play07:23

dikurang 0,4 atau = 0,1 mol nah ini

play07:28

kalau kita Ubah menjadi g Langsung kita

play07:30

kalikan saja dengan mr-nya mno2 adalah

play07:35

32 berarti 0,1 jika Lee 32 tuh sama

play07:40

dengan 3,2

play07:46

Hai jadi pereaksi yang bersisa adalah O2

play07:50

sisanya sebanyak 3,2 G jelas ya sekarang

play07:55

kita lanjut ke contoh yang kedua temen

play07:59

Hai sebanyak 11,2 gram logam Fe

play08:02

Diketahui f = 56 direaksikan dengan 0,4

play08:07

mol larutan H2 so4 menghasilkan fe2 so4

play08:11

3 kali dan gas hidrogen menurut

play08:15

persamaan reaksi V ditambah H2 so4

play08:18

menjadi fe2 so4 3 kali dan H2 volume gas

play08:23

hidrogen yang dihasilkan pada keadaan

play08:26

standar adalah kita buat persamaan

play08:29

reaksinya dulu ya yaitu V ditambah H2

play08:35

so4 menghasilkan fe2 so4 3 kali ditambah

play08:44

H2 lalu kita setarakan jumlah atom Fe di

play08:48

sebelah kiri satu di sebelah kanan dua

play08:51

berarti koefisien FS sebelah kiri kita

play08:53

ganti2 kemudian kita samakan jumlah atom

play08:57

S yang kita ikuti yang disebelah

play08:59

ya s-nya Ada tiga berarti di sebelah

play09:02

kiri harus tiga koefisien kita ganti 3

play09:05

lalu kita samakan jumlah atom O di

play09:08

sebelah kiri ada 12 3 kali 4 ya di

play09:11

sebelah kanan juga 12 berarti sudah sama

play09:14

terakhir kita samakan jumlah atom H kita

play09:17

ikuti Hai yang di sebelah kiri ada enam

play09:20

berarti di sebelah kanan juga harus 6

play09:22

maka koefisien kita ganti tiga

play09:25

berikutnya kita tulis data yang

play09:28

diketahui

play09:29

Hai sebanyak 11,2 gram logam Fe berarti

play09:33

efeknya Diketahui 11,2 g airnya 56 bisa

play09:41

kita langsung cari maunya ya jadi G

play09:44

dibagi dengan aritu akan menghasilkan

play09:47

Mal yaitu sebesar 0,2 mol

play09:53

hai lalu 0,4 mol larutan H2 so4 jadi H2

play09:59

so4 nya 0,4000560

play10:03

Hai yang diketahuinya sudah habis kita

play10:06

tentukan yang ditanyakan adalah volume

play10:09

gas hidrogen pada keadaan standar

play10:12

berarti H2 ya yang ditanya adalah vstp

play10:20

Hai untuk mencari volume H2 Artinya kita

play10:23

harus cari maunya sih A2 dulu

play10:25

menggunakan perbandingan koefisien dan

play10:28

perbandingan mol Nah di sini ada dua mal

play10:31

yang diketahui Mall mana yang menjadi

play10:33

acuan mal yang menjadi acuan adalah mol

play10:36

pereaksi pembatas ingat untuk menentukan

play10:40

reaksi pembatas caranya adalah

play10:42

masing-masing molt-pre aksi dibagi

play10:45

dengan koefisien masing-masing jadi 0,2

play10:49

dibagi2 0,4 dibagi3 yang nilainya lebih

play10:55

kecil itulah yang menjadi pereaksi

play10:58

pembatas kita hitung 0,2 dibagi dua

play11:02

adalah 0,1 fashya sedangkan 0,4 dibagi3

play11:07

lebih dari 0,1 berarti yang habis

play11:12

bereaksi adalah

play11:15

Hai atau disebut juga dengan pereaksi

play11:18

pembatas

play11:20

Hai nah yang menjadi acuan berarti

play11:22

mallnya sih segi-7 0,2 mau sekarang kita

play11:28

tentukan awalnya sih h20h 2 =

play11:34

koefisiennya H2 yaitu tiga dibagi dengan

play11:39

koefisien malpractice pembatas yaitu 2

play11:42

dikali dengan mallnya pereaksi pembatas

play11:46

yaitu 0,20 koma 2 dibagi dua 0,1 0,1

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dikali tiga samadengan 0,300

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Hai telah kita peroleh maunya sih A2

play12:00

Sekarang kita akan mencari volume STP

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dari CH2 ingat volume STP rumusnya

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adalah mal di kali 22,4 = 0,3 dikali

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22,4 atau sama dengan 6,72 l jadi volume

play12:26

gas hidrogen pada keadaan standar adalah

play12:29

6,72 l bisa dipahami ya Nah Sekian dulu

play12:34

untuk video kali ini terima kasih

play12:36

wassalamualaikum warahmatullahi

play12:38

wabarakatuh

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