Stoichiometry ( ) is the relationships between the quantities of reactants and products before, during and after chemical reactions.
Dalam ilmu kimia, stoikiometri () adalah ilmu yang mempelajari dan menghitung hubungan kuantitatif dari reaktan dan produk dalam reaksi kimia (persamaan kimia).
Thus, we get: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of liquid water.
Stoikiometri diilustrasikan melalui gambar berikut, dengan persamaan reaksi setara: CH4 + 2 O2 → CO2 + 2 H2O. Di sini, satu molekul metana bereaksi dengan dua molekul gas oksigen untuk menghasilkan satu molekul karbon dioksida dan dua molekul air.
Stoichiometry measures these quantitative relationships, and is used to determine the amount of products and reactants that are produced or needed in a given reaction.
Stoikiometri mengukur hubungan kuantitatif ini, dan digunakan untuk menentukan jumlah produk dan reaktan yang diproduksi atau dibutuhkan dalam reaksi yang diberikan.
Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry.
Menggambarkan hubungan kuantitatif antara zat-zat ketika mereka berpartisipasi dalam reaksi kimia dikenal sebagai stoikiometri reaksi.
This is called composition stoichiometry.
Hal ini disebut sebagai stoikiometri komposisi.
Gas stoichiometry deals with reactions solely involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases.
Stoikiometri gas berkaitan dengan reaksi yang melibatkan gas, di mana gas berada pada suhu, tekanan, dan volume yang diketahui dan dapat dianggap gas ideal.
For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products.
Untuk gas, rasio volume idealnya sama dengan hukum gas ideal, tetapi rasio massa dari reaksi tunggal harus dihitung dari massa molekul dari reaktan dan produk.
In practice, because of the existence of isotopes, molar masses are used instead in calculating the mass ratio.
Dalam praktiknya, karena keberadaan isotop, massa molar digunakan sebagai gantinya ketika menghitung rasio massa.
The term stoichiometry was first used by Jeremias Benjamin Richter in 1792 when the first volume of Richter's Anfangsgründe der Stöchyometrie oder Meßkunst chymischer Elemente (German for 'Fundamentals of Stoichiometry, or the Art of Measuring the Chemical Elements') was published.
Istilah stoikiometri pertama kali digunakan oleh Richter pada tahun 1792 ketika volume pertama Stoichiometry or the Art of Measuring the Chemical Elements karangan Richter diterbitkan.
Stoichiometry is not only used to balance chemical equations but also used in "conversions" between quantities of a substance by dimensional analysis, e.g., converting from grams to moles using molar mass as the "conversion factor", or from grams to milliliters using density.
Stoikiometri tidak hanya digunakan untuk menyeimbangkan persamaan kimia tetapi juga digunakan dalam konversi, misalnya, mengubah dari gram ke mol menggunakan massa molar sebagai faktor konversi, atau dari gram ke mililiter menggunakan kerapatan (densitas).
NaCl }}{58.44{\mbox{ g/mol}}}}=0.0342\ {\text{mol NaCl}}} In the above example, when written out in fraction form, the units of grams form a multiplicative identity, which is equivalent to one (g/g1), with the resulting amount in moles (the unit that was needed), as shown in the following equation, ( 2.00 g NaCl 1 ) ( 1 mol NaCl 58.44 g NaCl )
Dalam contoh di atas, ketika dituliskan dalam bentuk pecahan, satuan gram membentuk identitas multiplikatif, yang setara dengan satu (g/g1), dengan jumlah yang dihasilkan dalam mol (unit yang dibutuhkan), seperti yang ditunjukkan pada persamaan berikut, ( 2.00 g NaCl 1 ) ( 1 mol NaCl 58.44 g NaCl )
0.0342 mol NaCl }}{1}}\right)\left({\frac {1{\mbox{ mol NaCl}}}{58.44{\mbox{ g NaCl}}}}\right)=0.0342\ {\text{mol NaCl}}} Stoichiometry is often used to balance chemical equations (reaction stoichiometry).
0.034 mol }}{1}}\right)\left({\frac {1{\mbox{ mol NaCl}}}{58.44{\mbox{ g NaCl}}}}\right)=0.034\ {\text{mol}}} Stoikiometri sering digunakan untuk menyeimbangkan persamaan kimia (stoikiometri reaksi).
For example, the two diatomic gases, hydrogen and oxygen, can combine to form a liquid, water, in an exothermic reaction, as described by the following equation: 2 H2 + O2 → 2 H2O Reaction stoichiometry describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.
Sebagai contoh, dua gas diatomik, hidrogen dan oksigen, dapat bergabung untuk membentuk cairan, air, dalam reaksi eksotermik, seperti dijelaskan oleh persamaan berikut ini: 2 H2 + O2 → 2 H2O Stoikiometri reaksi menggambarkan perbandingan molekul hidrogen, oksigen, dan air 2: 1: 2 dalam persamaan di atas.
For example, in the reaction 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O the amount of water that will be produced by the combustion of 0.27 moles of CH3OH is obtained using the molar ratio between CH3OH and H2O of 2 to 4. ( 0.27 mol C H 3 O H 1 ) ( 4 mol H 2 O 2 mol C H 3 O H )
Misalnya dalam reaksi 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O jumlah air yang akan dihasilkan oleh pembakaran 0.27 mol CH3OH diperoleh dengan menggunakan rasio molar antara CH3OH dan H2O dari 2 menjadi 4. ( 0.27 mol C H 3 O H 1 ) ( 4 mol H 2 O 2 mol C H 3 O H )
0.54 mol H 2 O }\mathrm {CH_{3}OH} }{1}}\right)\left({\frac {4{\mbox{ mol }}\mathrm {H_{2}O} }{2{\mbox{ mol }}\mathrm {CH_{3}OH} }}\right)=0.54\ {\text{mol }}\mathrm {H_{2}O} } The term stoichiometry is also often used for the molar proportions of elements in stoichiometric compounds (composition stoichiometry).
0.54 mol H 2 O }\mathrm {CH_{3}OH} }{1}}\right)\left({\frac {4{\mbox{ mol }}\mathrm {H_{2}O} }{2{\mbox{ mol }}\mathrm {CH_{3}OH} }}\right)=0.54\ {\text{mol }}\mathrm {H_{2}O} } Istilah stoikiometri juga sering digunakan untuk proporsi molar unsur-unsur dalam senyawa stoikiometris (stoikiometri komposisi).
For example, the stoichiometry of hydrogen and oxygen in H2O is 2:1.
Misalnya, stoikiometri hidrogen dan oksigen dalam H2O adalah 2:1.
In stoichiometric compounds, the molar proportions are whole numbers.
Dalam senyawa stoikiometris, the molar proportions are whole numbers.
54.32 g Ag }}{1}}\right)\left({\frac {107.87{\mbox{ g Ag}}}{1{\mbox{ mol Ag}}}}\right)=54.32\ {\text{g Ag}}} This set of calculations can be further condensed into a single step: m A g( 16.00 g C u 1 ) ( 1 mol C u 63.55 g C u ) ( 2 mol A g 1 mol C u ) ( 107.87 g A g 1 mol Ag )
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} Zumdahl, Steven S. Chemical Principles.
54.32 g }=\left({\frac {16.00{\mbox{ g }}\mathrm {Cu} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Cu} }{63.55{\mbox{ g }}\mathrm {Cu} }}\right)\left({\frac {2{\mbox{ mol }}\mathrm {Ag} }{1{\mbox{ mol }}\mathrm {Cu} }}\right)\left({\frac {107.87{\mbox{ g }}\mathrm {Ag} }{1{\mbox{ mol Ag}}}}\right)=54.32{\mbox{ g}}} For propane (C3H8) reacting with oxygen gas (O2), the balanced chemical equation is: C3H8 + 5 O2 → 3 CO2 + 4 H2O The mass of water formed if 120 g of propane (C3H8) is burned in excess oxygen is then m H 2 O( 120.
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} Zumdahl, Steven S. Chemical Principles.
196 g O} }=\left({\frac {120.{\mbox{ g }}\mathrm {C_{3}H_{8}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {C_{3}H_{8}} }{44.09{\mbox{ g }}\mathrm {C_{3}H_{8}} }}\right)\left({\frac {4{\mbox{ mol }}\mathrm {H_{2}O} }{1{\mbox{ mol }}\mathrm {C_{3}H_{8}} }}\right)\left({\frac {18.02{\mbox{ g }}\mathrm {H_{2}O} }{1{\mbox{ mol }}\mathrm {H_{2}O} }}\right)=196{\mbox{ g}}}
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} Zumdahl, Steven S. Chemical Principles.
Stoichiometry is also used to find the right amount of one reactant to "completely" react with the other reactant in a chemical reaction – that is, the stoichiometric amounts that would result in no leftover reactants when the reaction takes place.
Stoikiometri juga digunakan untuk menemukan jumlah yang tepat dari satu reaktan untuk "sepenuhnya" bereaksi dengan reaktan lain dalam reaksi kimia–yaitu, jumlah stoikiometris yang akan menghasilkan tidak ada reaktan sisa ketika reaksi berlangsung.
So, to completely react with 85.0 g of iron(III) oxide (0.532 mol), 28.7 g (1.06 mol) of aluminium are needed.
Maka untuk tepat mereaksikan 85.0 g besi(III) oksida (0.532 mol), 28.7 g (1.06 mol) aluminium dibutuhkan.
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed when the reaction is complete.
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} Zumdahl, Steven S. Chemical Principles.
186.6 g }=\left({\frac {200.0{\mbox{ g }}\mathrm {PbS} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {PbS} }{239.27{\mbox{ g }}\mathrm {PbS} }}\right)\left({\frac {2{\mbox{ mol }}\mathrm {PbO} }{2{\mbox{ mol }}\mathrm {PbS} }}\right)\left({\frac {223.2{\mbox{ g }}\mathrm {PbO} }{1{\mbox{ mol }}\mathrm {PbO} }}\right)=186.6{\mbox{ g}}} m P b O( 200.0 g O 2 1 ) ( 1 mol O 2 32.00 g O 2 ) ( 2 mol P b O 3 mol O 2 ) ( 223.2 g P b O 1 mol P b O )
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} Zumdahl, Steven S. Chemical Principles.
930.0 g }=\left({\frac {200.0{\mbox{ g }}\mathrm {O_{2}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {O_{2}} }{32.00{\mbox{ g }}\mathrm {O_{2}} }}\right)\left({\frac {2{\mbox{ mol }}\mathrm {PbO} }{3{\mbox{ mol }}\mathrm {O_{2}} }}\right)\left({\frac {223.2{\mbox{ g }}\mathrm {PbO} }{1{\mbox{ mol }}\mathrm {PbO} }}\right)=930.0{\mbox{ g}}} Because a lesser amount of PbO is produced for the 200.0 g of PbS, it is clear that PbS is the limiting reagent.
28.7 g }=\left({\frac {85.0{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }{1}}\right)\left({\frac {1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }{159.7{\mbox{ g }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {2{\mbox{ mol Al}}}{1{\mbox{ mol }}\mathrm {Fe_{2}O_{3}} }}\right)\left({\frac {26.98{\mbox{ g Al}}}{1{\mbox{ mol Al}}}}\right)=28.7{\mbox{ g}}} Zumdahl, Steven S. Chemical Principles.
Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known.
Stoikiometri gas berkaitan dengan reaksi yang melibatkan gas, di mana gas berada pada suhu, tekanan, dan volume yang diketahui dan dapat dianggap gas ideal.
Internal Combustion Engine Fundamentals, John B. Heywood Engine Combustion primer from the University of Plymouth Free Stoichiometry Tutorials from Carnegie Mellon's ChemCollective Stoichiometry Add-In for Microsoft Excel Archived 2011-05-11 at the Wayback Machine for calculation of molecular weights, reaction coëfficients and stoichiometry.
Internal Combustion Engine Fundamentals, John B. Heywood (Inggris) Engine Combustion primer dari Universitas Plymouth (Inggris) Free Stoichiometry Tutorials dari Carnegie Mellon's ChemCollective (Inggris) Stoichiometry Add-In for Microsoft Excel Diarsipkan 2011-05-11 di Wayback Machine.
Reaction Stoichiometry Calculator a comprehensive free online reaction stoichiometry calculator.
(Inggris) Reaction Stoichiometry Calculator kalkulator stoikiometri reaksi daring gratis yang komprehensif.