Как составить электронный баланс химия онлайн

Электронный учебник

Инструкция по балансировке химических уравнений: Введите уравнение химической реакции и нажмите “Уравнять”. Ответ на этот вопрос появится ниже Всегда используйте верхний регистр для первого символа в названии химического элемента и нижнем регистре для второго символа. Например: Fe, Au, Co, C, O, N, F. Сравните: Co – кобальт и CO – угарный газ Для уравнивания полуреакции окислительно-восстановительного процесса используйте {-} или е Для обозначения зарядов ионов используйте фигурные скобки: {+3} или {3+} или {3}. Пример: Fe {3 +} +. I {-} = Fe {2 +} + I2 В случае сложных соединений с повторяющимися группами, замените неизменные части в формуле реагентов. Например, уравнение C6H5C2H5 + O2 = C6H5OH + CO2 + H2O не будет сбалансированно, но если C6H5 заменить на X, то все получится PhC2H5 + O2 = PhOH + CO2 + H2O  Примеры полных уравнений химического баланса: Fe + Cl2 = FeCl3 KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2 K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO C6H5COOH + O2 = CO2 + H2O K4Fe(CN)6 + KMnO4 + H2SO4 = KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O Cr2O7{-2} + H{+} + {-} = Cr{+3} + H2O S{-2} + I2 = I{-} + S PhCH3 + KMnO4 + H2SO4 = PhCOOH + K2SO4 + MnSO4 + H2O CuSO4*5H2O = CuSO4 + H2O calcium hydroxide + carbon dioxide = calcium carbonate + water sulfur + ozone = sulfur dioxide  Примеры химических уравнений реагентов (полное уравнение будет предложено): H2SO4 + K4Fe(CN)6 + KMnO4 Ca(OH)2 + H3PO4 Na2S2O3 + I2 C8H18 + O2 hydrogen + oxygen propane + oxygen  Related chemical tools: Вычисление молярной массы pH solver

Химические уравнения, сбалансированные сегодня

Оставьте нам отзыв о своем опыте работы с химической балансировкой уравнений.

Химические Уравнения онлайн!

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Или Попробуйте случайную реакцию

Окислительно-восстановительные реакции, также редокс (англ. redox, от reduction-oxidation — восстановление-окисление) — это встречно-параллельные химические реакции, протекающие с изменением степеней окисления атомов, входящих в состав реагирующих веществ (или ионов веществ), реализующихся путём перераспределения электронов между атомом-окислителем (акцептором) и атомом-восстановителем (донором).

Калькулятор сбалансирования окислительно-восстановительной реакции

Онлайн калькулятор для уравнивания(сбалансирования) несбалансированного окислительно-восстановительной химической реакции.

Описание окислительно-востановительной реакции

В процессе окислительно-восстановительной реакции восстановитель отдаёт электроны, то есть окисляется; окислитель присоединяет электроны, то есть восстанавливается. Причём любая окислительно-восстановительная реакция представляет собой единство двух противоположных превращений — окисления и восстановления, происходящих одновременно и без отрыва одного от другого

Пример окислительно-востановительной реакции

Методом электронного баланса подберите коэффициенты в схемах следующих окислительно-восстановительных реакций с участием металлов:

а) Ag + HNO3 → AgNO3 + NO + H2O
б) Ca +H2SO4 → CaSO4 + H2S + H2O
в) Be + HNO3 → Be(NO3)2 + NO + H2O

Применение метода электронного баланса по шагам. Пример «а»

Составим электронный баланс для каждого элемента реакции окисления Ag + HNO₃ → AgNO₃ + NO + H₂O.

Шаг 1. Подсчитаем степени окисления для каждого элемента, входящего в химическую реакцию.

Ag. Серебро изначально нейтрально, то есть имеет степень окисления ноль.

Для HNO₃ определим степень окисления, как сумму степеней окисления каждого из элементов.

Степень окисления водорода +1, кислорода -2, следовательно, степень окисления азота равна:

0 — (+1) — (-2)*3 = +5

(в сумме, опять же, получим ноль, как и должно быть)  

Теперь перейдем ко второй части уравнения.

Для AgNO₃ степень окисления серебра +1 кислорода -2, следовательно степень окисления азота равна:

0 — (+1) — (-2)*3 = +5

Для NO степень окисления кислорода -2, следовательно азота +2

Для H₂O степень окисления водорода +1, кислорода -2

Шаг 2. Запишем уравнение в новом виде, с указанием  степени окисления каждого из элементов, участвующих в химической реакции.

Ag⁰ + H⁺¹N⁺⁵O^-2₃ → Ag⁺¹N⁺⁵O^-2₃ + N⁺²O^-2 + H⁺¹₂O^-2

Из полученного уравнения с указанными степенями окисления, мы видим несбалансированность по сумме положительных и отрицательных степеней окисленияотдельных элементов.

Шаг 3. Запишем их отдельно в виде электронного баланса — какой элемент и сколько теряет или приобретает электронов:
(Необходимо принять во внимание, что элементы, степень окисления которых не изменилась — в данном расчете не участвуют)

Ag⁰ — 1e = Ag⁺¹
N⁺⁵ +3e = N⁺²

Серебро теряет один электрон, азот приобретает три. Таким образом, мы видим, что для балансировки нужно применить коэффициент 3 для серебра и 1 для азота. Тогда число теряемых и приобретаемых электронов сравняется.

Шаг 4. Теперь на основании полученного коэффициента «3» для серебра, начинаем балансировать все уравнение с учетом количества атомов, участвующих в химической реакции.

• В первоначальном уравнении перед Ag ставим тройку, что потребует такого же коэффициента перед AgNO₃
• Теперь у нас возник дисбаланс по количеству атомов азота. В правой части их четыре, в левой — один. Поэтому ставим перед HNO₃ коэффициент 4
• Теперь остается уравнять 4 атома водорода слева и два — справа. Решаем это путем применения коэффииента 2 перед H₂O

Ответ:  3Ag + 4HNO₃ = 3AgNO₃ + NO + 2H₂O

Пример «б»

Составим электронный баланс для каждого элемента реакции окисления Ca +H₂SO₄ → CaSO₄ + H₂S + H₂O

Для H₂SO₄  степень окисления водорода +1 кислорода -2 откуда степень окисления серы 0 — (+1)*2 — (-2)*4 = +6

Для CaSO₄  степень окисления кальция равна +2 кислорода -2 откуда степень окисления серы 0 — (+2) — (-2)*4 = +6

Для H₂S степень окисления водорода +1, соответственно серы -2

Ca⁰ +H⁺¹₂S⁺⁶O^-2₄ → Ca⁺²S⁺⁶O^-2₄ + H⁺¹₂S^-2 + H⁺¹₂O^-2
Ca⁰ — 2e = Ca⁺² (коэффициент 4)
S⁺⁶ + 8e = S^-2

4Ca + 5H₂SO₄ = 4CaSO₄ + H₂S + 4H₂O

Пример «в»

Составим электронный баланс для каждого элемента реакции окисления Be + HNO₃ → Be(NO₃)₂ + NO + H₂O

HNO₃ см. выше

Для Be(NO₃)₂ степень окисления бериллия +2, кислорода -2, откуда степень окисления азота ( 0 — (+2) — (-2)*3*2 ) / 2 = +5

NO см. выше

H₂O см. выше

Be⁰ + H⁺¹N⁺⁵O^-2₃ → Be⁺²(N⁺⁵O^-2₃)₂ +  N⁺²O^-2 +  H⁺¹₂O^-2
Be⁰ — 2e = Be⁺² (коэффициент 3)
N⁺⁵ +3e = N⁺² (коэффициент 2)

3Be + 8HNO₃ → 3Be(NO₃)₂ + 2NO + 4H₂O

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How To Balance Chemical Equations

Last Updated: April 13, 2023

In a chemical reaction, the quantity of each element does not change, meaning both sides of the equation must represent the same quantity of each element or particle (if they didn’t, it would violate the law of conservation of mass).
In case of net ionic reactions, the same charge must be present on both the reactant side and product side of the unbalanced equation. By changing the coefficients for each compound, a chemical equation can be balanced.
There are three common methods for balancing chemical equations: inspection, linear systems, or using a chemical equation balancer (such as the one on this page).

• Balance Using Inspection
• Balance Using Linear Systems
• Balance Using Inspection + Linear Systems
• Balance Using An Online Balancer
• Popular Chemical Equations to Balance
• Questions about Balancing Equations

Using Inspection/Trial and Error

The easiest way for beginners to balance simple chemical equations by hand is via inspection (also called trial and error).
The basic idea is to balance one element at a time (usually starting with the most complicated molecule, and ending with hydrogen and oxygen) until all the elements are balanced.

Example #1 (Trivial): CaCO3 + HCl

To start off with a simple example, lets balance the acid-base reaction of two salts, Calcium Carbonate (CaCO3) + Hydrochloric Acid (HCl) → Calcium Chloride (CaCl2) + Carbon Dioxide (CO2) + Water (H2O):

CaCO₃ + HCl = CaCl₂ + CO₂ + H₂O

If we count up the number of each element on both the left-hand-side and right-hand-side, we see that all but Hydrogen and Chlorine are balanced.
There is 1 H and 1 Cl on the left, but 2 of each the right.
To balance H and Cl, we can put a 2 in front of HCl on the left-hand-side:

CaCO₃ + 2 HCl = CaCl₂ + CO₂ + H₂O

Now that there is an equal quantity of Ca, C, Cl, H and O on both sides, the chemical equation is balanced.

Even though it was simple, there are actually quite a few cases where you can balance the chemical equation in one step:

• Iron + Hydrogen Chloride = Ferrous Chloride + Hydrogen Gas: Fe + HCl → FeCl2 + H2
• Zinc + Hydrogen Chloride = Zinc Chloride + Hydrogen Gas: Zn + HCl → ZnCl2 + H2

Also, be aware that sometimes no balancing is needed. That is the case in chemical equations like:

• CaO + H2O → Ca(OH)2

Example #2 (Simple): Na + H2O

The previous example was simpler than most but demonstrated the basic concepts.
Usually, balancing chemical equations will require multiple steps.
Take for example the exothermic reaction of Sodium (Na) and Water (H2O), which releases heat, Sodium Hydroxide (NaOH) and Hydrogen Gas (H2):

Na + H₂O = NaOH + H₂

In this case, there are an equal number of Na and O atoms, but like last time, H needs to be balanced, with 2 on the left, and 3 on the right.
Lets start by balancing hydrogen. Because 3 is not divisible by 2, it means coefficients of multiple compounds containing H will need to be changed.
First, lets put a 2 in front of H₂O:

Na + 2 H₂O = NaOH + H₂

Notice how the 2 in front of H2O is “distributed” to both the H₂ and the O. We now have 4 H, 2 O and 1 Na atom on the left, but 3 H, 1 O and 1 Na atom on the right.
We went from 1 unbalanced element to multiple. This is common and doesn’t mean any mistakes were made.

As was mentioned before, we knew we had to update multiple coefficients to balance the hydrogen atoms. As a next step, lets balance H again by putting a 2 in front of NaOH so the equation reads:

Na + 2 H₂O = 2 NaOH + H₂

You might wonder why we didn’t instead increase the number of H2 molecules. We could have tried, but we would have realized that we would need to balance H again (hence the trial and error).

If we count up the number of atoms again, we have 4 H and 2 O on both sides, but there is 1 sodium atom on the left side, and 2 sodium atoms on the right side.
In the previous step, we actually ended up balancing the oxygen atoms as well!
But sadly, this is still an unbalanced equation 😢. Lets fix that by putting a 2 in front of Na:

2 Na + 2 H₂O = 2 NaOH + H₂

And there you have it! After you count everything up, you’ll see there is an equal amount of each chemical on both sides, so our work is done!👍

Try practicing what you’ve learned so far on:

• Combustion of Methane and Oxygen: CH4 + O2 = CO2 + H2O

Example #3 (Complex): Cu + HNO3

Now that we’ve hopefully gotten the hang of balancing simple chemical equations, lets move onto something a bit more complex, such as the reaction of Copper (Cu) and Nitric Acid (HNO3) into Copper(II) Nitrate (Cu(NO3)2) + Nitrogen Dioxide (NO2) + Water (H2O):

Cu + HNO₃ = Cu(NO₃)₂ + NO₂ + H₂O

The left hand side has 1 Cu and the right hand side has 1 Cu, so the Cu atoms are balanced.
However, the left hand side has 1 H, 1 N, and 3 O’s and the right hand side has 2 H’s, 3 N’s and 9 O’s.
This equation is not balanced because there are an unequal amount of H’s, N’s and O’s on both sides of the equation.

To balance the given equation, put a 3 in front of the HNO₃ on the left hand side to get 3 N’s on both sides:

Cu + HNO₃ = 3 Cu(NO₃)₂ + NO₂ + H₂O

As a side-effect, O was also balanced with 9 on each side.

Since H is the only remaining unbalanced element, in the next step, we will put a 2 in front of H2O to try to balance it.
Hopefully you can see that O will become unbalanced as a result, and since like before, 3 is not divisible by 2, multiple steps to balance hydrogen will be needed, so there is more work ahead, but we’ve got this 💪:

Cu + HNO₃ = 3 Cu(NO₃)₂ + NO₂ + 2 H₂O

Tallying things up again, we have:

	Cu + 3HNO3	Cu(NO3)2 + NO2 + 2H2O
Cu	1	1
H	3	4
N	3	3
O	9	10

For more complex equations, keeping a table like this can make it easier to keep track of your progress. As you change coefficients, update the table with the latest atom counts.

As we predicted, and as can be seen in the table above, we still need to balance H and O.
To do that, lets add one more molecule of HNO3 to get:

Cu + HNO₃ = 4 Cu(NO₃)₂ + NO₂ + 2 H₂O

	Cu + 4HNO3	Cu(NO3)2 + NO2 + 2H2O
Cu	1	1
H	4	4
N	4	3
O	12	10

Even though we’ve now balanced H, we need to balance N again 😔. Don’t be discouraged though, we’re actually quite close!
If you look closely, you’ll notice we just need 1 N and 2 O’s on the right to make things balanced, and luckily enough, we have a NO2 molecule on the right. If we add one more, we get:

Cu + HNO₃ = 4 Cu(NO₃)₂ + 2 NO₂ + 2 H₂O

The equation is now balanced because there is an equal amount of substances on the left-hand side and the right-hand side of the equation.

Example #4 (Complex): Fe + HCl

As the final example, lets balance the single-displacement redox reaction of Iron + Hydrochloric Acid = Ferric Chloride + Hydrogen Gas: Fe + HCl → FeCl3 + H2:

Fe + HCl → FeCl₃ + H₂

While the equation may seem simple, this equation is more complex than the previous examples and requires more steps so lets start by writing out the table to keep track of each type of atom:

	Fe + HCl	FeCl3 + H2
Fe	1	1
H	1	2
Cl	1	3

In this case, Fe is balanced, but H and Cl are not. Following our suggestions from before to balance O and H last and more complicated molecules (FeCl3 in this example) first (while not necessary, it usually makes things simpler), lets start by balancing Cl by puting a 3 in front of HCl.

Fe + 3 HCl → FeCl₃ + H₂

	Fe + 3HCl	FeCl3 + H2
Fe	1	1
H	3	2
Cl	3	3

Next, lets balance H. Like before, since 3 isn’t divisible by 2, it will require multiple steps, but unlike last time, to save time, lets balance H in one go using some basic math.
Since H only exists in one substance on each side, lets find the lowest common multiple of 3 and 2, which is 6, and then increase coefficients to each side to make both sides have 6 hydrogen atoms:

Fe + 6 HCl → FeCl₃ + 3 H₂

	Fe + 3HCl	FeCl3 + 3H2
Fe	1	1
H	6	6
Cl	6	3

Again, the table makes things look deceptively close to being done, but we still have a couple of steps remaining to balance the equation. To balance Cl, lets add one more FeCl2 molecule:

Fe + 6 HCl → 2 FeCl₃ + 3 H₂

	Fe + 3HCl	2FeCl3 + 3H2
Fe	1	2
H	6	6
Cl	6	6

Now Iron is unbalanced (for the first time), but we have an easy fix. Add one more Fe atom on the left and we get:

2 Fe + 6 HCl → 2 FeCl₃ + 3 H₂

	2Fe + 3HCl	2FeCl3 + 3H2
Fe	2	2
H	6	6
Cl	6	6

While it took us a few tries, we finally managed to balance the equation.

Practice Balancing Chemical Equations Using Inspection

Hopefully by now you’ve mastered the inspection / trial and error method for balancing chemical equations.
Practice by balancing a few of the equations below.
If you get stuck, click the links to use our chemical equation balance calculator to see the balanced result and the four easy steps to get there:

• Aluminium + Sodium Hydroxide + Water = Sodium Aluminate + Hydrogen Gas: Al + NaOH + H2O = NaAlO2 + H2
• Zinc + Hydrogen Chloride = Zinc Chloride + H2 Gas: Zn + HCl → ZnCl2 + H2

Using Linear Systems of Equations

In reactions involving many compounds or elements (usually both!), it may not be easy or practical to balance the equations using inspection.
Instead, the equations can be balanced using the algebraic method, based on solving a set of linear equations.
Hopefully by now you’re at least somewhat familiar with at least basic algebra, or to make things even easier, linear algebra.
Linear systems, while seemingly complex can easily balance even hard chemical equations, and can be made easier by using a calculator to solve the systems of equations.
If using matrices to solve the linear equations, this is also sometimes called the matrix method.

Example: Balance KMnO4 + HCl Using The Algebraic Method

To start, write out the unbalanced equation for Potassium Permanganate (KMnO4) + Hydrogen Chloride (HCl) = Potassium Chloride (KCl) + Manganese(II) Chloride (MnCl2) + Dichlorine (Cl2) + Water (H2O):

KMnO4 + HCl = KCl + MnCl2 + Cl2 + H2O

The coefficients in balanced chemical equations represent both the basic unit and the mole ratios of the substances.
As the first step we assign variables for each of the unknown coefficients. In this example we will use a–f but you can use any names you prefer.
Tip: If your equation contains lone electrons denoted by e, you may want to skip e as a variable to avoid confusion 😂.

a KMnO₄ + b HCl = c KCl + d MnCl₂ + e Cl₂ + f H₂O

Next, we will create a system of linear equations, with each equation representing one of the elements in the equation, and each term in the equations representing the total number of atoms of each element in the corresponding substance.
Since there must be the same quantities of each atom on each side of the equation due to the conservation of mass, we will set the reactant terms equal to the product terms.
If all that sounds complicated, it will make more sense once we write out the equations:

Equation: KMnO4 + HCl = KCl + MnCl2 + Cl2 + H2O
       K: 1a          = 1c
      Mn: 1a          =       1d
       O: 4a          =                     1f
       H:         1b  =                     2f
      Cl:         1b  = 1c  + 2d    + 2e

Overall, this is a relatively simple example other than the equation for Cl.
We have 5 equations and 6 unknowns, which means we technically have infinite solutions, but in this case since we only have 1 more unknown than equations, we can just take the simplest whole number solution.
You can use whatever method you know for solving systems of linear equations (e.g. substitution, elimination, reduced row echelon forms, gauss-jordan elimination, or a calculator) to get through to the next step of solving for each unknown:

• a = ¹/₄ f
• b = 2 f
• c = ¹/₄ f
• d = ¹/₄ f
• e = ⁵/₈ f
• f ∈ R

Since we want the result to have the smallest integer coefficients, f needs to be the lowest common multiple of the denominators of each coefficient in the other equations, which in this case is 8.
Substitute 8 for f to get:

• a = 2
• b = 16
• c = 2
• d = 2
• e = 5
• f = 8

Finally, substitute the known values into the original equation with variable coefficients to get the balanced equation:

2 KMnO₄ + 16 HCl = 2 KCl + 2 MnCl₂ + 5 Cl₂ + 8 H₂O

Practice Balancing Chemical Equations Using Linear Systems

Try practicing the method described earlier by balancing these equations:

• K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO

If you get stuck, click on the equation to get a step-by-step solution using the algebraic method using our chemical equation balancer.

Combining Inspection + Linear Systems

To make the process faster and easier, inspection and linear systems can be combined to get a more practical algorithm:

1. Identify elements which occur in one compound on each side. This is very common (e.g. K, Mn, O and H in the prior example).
2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, such as a. In the prior example, O has the largest index (4)
   • a KMnO4 + HCl = KCl + MnCl2 + Cl2 + H2O
3. Next, start determining other coefficients in terms of a: H2O has to be 4a (because of O), MnCl2 has to be 1a (because of Mn), and KCl has to be 1a (because of K).
   This eliminates the first three equations of the system, leaving just two. It is already known that whatever the coefficients are, these ratios must hold:
   • a KMnO4 + HCl = 1a KCl + 1a MnCl2 + Cl2 + 4a H2O
4. One can continue by writing the equations now (and having a simpler problem to solve) or, in this particular case you could notice that to balance the Hydrogens between HCl and 4a H2O, you need 8a HCls,
   and then, to balance the Chlorine atoms between 8a HCl and Cl2, you need 5/2 a (or 2.5a) Cl2 (note the 1a Cl in of KCl and 2a of Cl in MnCl2, so the equation to solve is 8a Cl = 1a Cl + 1a Cl2 + ? Cl2).
   • a KMnO4 + 8a HCl = 1a KCl + 1a MnCl2 + 2.5a Cl2 + 4a H2O
5. After finding a convenient value for a (in this case since 5/2 is the only fraction, 2 will work to cancel the denominator) the result is:
   • 2 KMnO4 + 16 HCl = 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H2O

As you can see, this was the same result as we got earlier using the algebraic method for balancing chemical equations.

Using a Chemical Equation Balancer Online Tool

The simplest and fastest method to balance chemical equations by far is to use an online chemical equation balancer.
The ChemicalAid balancing equations calculator is the most advanced chemical balancer on the web.
Our chemical equation balancer will find the coefficients to balance the chemical equation, determine the type of reaction the equation represents, its word equation, the thermodynamics of the equation, and provide the steps to balance it using both the inspection and algebraic methods we covered earlier.
You can also use the calculator to double-check your work and perform follow-up stoichiometry and limiting reagent calculations.
When using online equation balancing apps, be sure to use the proper capitalization for each element.

Note that the two methods described earlier to balance equations aren’t the only ones.
Redox (Reduction-Oxidation) chemical equations should be balanced using the ion-electron method, oxidation number method or ARS method.
For these, you can use our redox calculator to ensure the redox reactions are balanced according to their oxidation states. If you need to find a net ionic equation, use our ionic reaction calculator.

Popular Chemical Equations

It’s common that you are given an equation to balance where you only know the reactant or the product chemicals, not both.
In these cases, you need to predict the products or reactants (tip: our balancing chemical equations calculator can usually work as a chemical equation product calculator if just the reactants are entered).
To help we’ve compiled a list of common reactions to balance:

• cu+hno3
• nh3+o2
• al+hcl
• co2+h2o
• caco3+hcl
• na2o + h2o
• mg + hcl
• zn+h2so4
• ca+h2o
• al+h2so4
• naoh + hcl
• p2o5+h2o
• na2co3+hcl
• s+o2
• k+h2o
• cac2 + h2o
• p+o2
• naoh+h2so4
• fe + h2so4
• mg + hno3
• al + hno3
• so2+h2o
• al+o2
• cu+h2so4
• h2s+o2
• Sodium Chloride (Table Salt) Dissolves in Water: NaCl + H2O = Na⁺(aq) + Cl^–(aq)
• kmno4 + hcl
• so3+h2o
• nahco3 + hcl
• mg+o2
• al2o3 + naoh
• ch3cooh + naoh
• fe+o2
• naoh+co2
• mg+h2so4
• h2+o2
• cu+hcl
• na+o2
• c+o2
• cuso4+naoh
• nh3+hcl
• ca(oh)2 + co2
• fe+cl2
• n2+o2
• so2+o2
• nh3+h2o
• fe + hno3

Frequently Asked Questions About Balancing Chemical Equations

• How do you write chemical equations?
  • When writing chemical equations on a computer (such as when inputting an equation into a chemical balancer), it is important to pay attention to details and follow these rules:
    • Element symbols in each chemical formula must be properly capitalized. The first letter of each element should be capitalized, and any following must be lowercase. Use a periodic table to ensure you are entering the correct symbols.
    • Stoichiometric coefficients go before each substance (if omitted, it means 1).
    • Net charges should be in superscript after each substance, along with a “+” or “-“, but before the state (if omitted, it means 0).
    • States of matter (s, l, g, aq) should be all lower-case in parentheses after each substance.
    • If there are multiple reactants or multiple products, they can be separated with a plus sign (“+”).
    • The reactants and products are separated with an arrow (note that different arrows can represent different types of reactions).
• Why do chemical equations need to be balanced?
  • The law of conservation of mass states that matter cannot be created or destroyed.
    This means that in a reaction, the reactants (left-hand side) and products (right-hand side) must have an equal number of atoms of each element.
    The purpose of balancing chemical equations is to ensure this is the case.
• How do you balance an equation with fractions?
  • Chemical equations generally shouldn’t have fractions in them, but in some rare cases they may be written as such.
    This may also be the case during the process of trying to balance an equation.
    If your chemical equation has fractions in it, convert all numbers into improper fractions, find the lowest common multiple (LCM) of all denominators, and multiply each number by the LCM.
• How do you identify the type of reaction based on the chemical equation?
  • Our chemical equation balancer will automatically calculate the theoretical equation type based on the balanced chemical equation.
    You can read our article on the different types of chemical reactions to find out how.