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If you are struggling to balance chemical equations as a student of chemistry, there is nothing to worry about, because we have got you covered. Our equation balancer offers to balance your chemical equations on one click. You don't have to rearrange the number of atoms manually for any equation from right now. We will discuss how you can use balancing chemical equations calculator and how to balance a chemical equation manually with a complete process.
How to use balance equation calculator?
Balancing equations calculator has a simple to use interface. It is completely free of cost, and everyone can use it on the go. For your ease and comfort, we have provided an interactive periodic table. You can click on any element, and it will automatically include that element in your equation. You can also use number digits to specify the number of atoms for each element in your equation. You don't have to write anything because everything can be done with simple clicks using this balanced equation calculator.
To use the chemical equation calculator, simply make your equations by clicking on the elements in the above periodic table. Or you can write your equation in the input box. Enter the chemical equation that is unbalanced because a chemical balancer will show an error message if the given equation is already in a balanced form. After entering your equation in the input box, click the "Calculate" button. The chemical reaction balancer will give you the balanced equation. It will highlight the digits in blue color that it has added to balance the equation. You can reset your equation by clicking the "Reset" button.
How to balance chemical equations?
The quantity of each element does not change in a chemical reaction. Each side of the equation must, therefore, represent the same amount of each specific element. The same charge will exist on both sides of the unbalanced equation in case of net ionic reactions. The equation can be balanced by increasing the scalar number for every molecular formula. We will use two methods to balance chemical equations here.
Trial and Error/Inspection method?
We will balance the simple and complex equations with each method so that you can deeply understand the process of balancing chemical equations. Trial and Error/Inspection method is used to balance simple chemical equations. The most complex molecule will usually be balanced first. Typically, oxygen and hydrogen are balanced at the end.
Simple Equation
$\mathrm{Na + O_2 = Na_2 O}$
To balance the above equation, which includes sodium and oxygen, this chemical equation must have the same amount of Na on the right side as well as on the left side. As it is now, on the left is 1 Na, but on the right, there are 2 Na. The solution to this problem is to put a 2 on the left side of the Na.
$\mathrm{2 Na + O_2 = Na_2 O}$
Now, there is a total of 2 Na on both sides of this equation. But oxygen atoms are not balanced yet. We will balance the atoms of oxygen now. There is only one oxygen atom on the right side of the equation and 2 on the left side. This chemical equation is still unbalanced. To make oxygen atoms equal, 2 is inserted on the right side of the equation with Na2O. The equation will be as follow after adding 2 on the right side of the equation:
$\mathrm{2 Na + O_2 = 2 Na_2O}$
Notice that the 2 is applicable to both O and Na2 on the right side. The left side of the equation currently consists of 2 Na atoms and 2 O atoms. There are 4 Na's and 2 O's on the right side of the equation. Again, it is not balanced yet; every atom on both sides must be identical in weight. 2 more Na's are added on the left side of the equation to fix this issue. Now the equation will be like:
$\mathrm{4 Na + O_2 = 2 Na_2 O}$
This equation is a balanced equation because, on the left and right sides of the equation, there are equal numbers of atoms after balancing.
Complex Equation
$\mathrm{P_4 + O_2 = 2 P_2 O_5}$
The above equation is an unbalanced complex chemical equation because, on both sides of the equation, O exists in different numbers. On the left side, there are 4 P, and on the right side, there are also 4 P, it means P is equal on both sides. There are 2 O on the left side and 10 O on the right side.
Add 5 with O2 on the left side of this chemical formula, which will make 10 O on both sides of the equation. After adding 5 on the left side, the equation will look like:
$\mathrm{P_4 + 5 O_2 = 2 P_2 O_5}$
The equation is now balanced, as on the left and right side of the equation, there are equal amounts of substances for each element.
One more complex equation with this method
$\mathrm{C_2 H_5 OH + O_2 = C O_2 + H_2 O}$
This calculation is complicated and requires more steps than the previous examples. The more complicated molecule here is $\mathrm{C_2 H_5 OH}$, and to balance carbon atoms, start by placing coefficient 2 before $\mathrm{CO_2}$.
\(\mathrm{C_2 H_5 OH + O_2 = 2 C O_2 + H_2 O}\)
Since C 2 H 5 OH comprises of six hydrogen atoms on the left side, it is possible to balance hydrogen atoms on the right side by adding 3 with \(\mathrm{H_2O}\).
\(\mathrm{C_2 H_5 OH + O_2 = 2 C O_2 + 3 H_2 O}\)
Now we will balance the oxygen atoms in this equation. Since there are only 3 oxygen atoms on the left side and 7 oxygen atoms to the right, we will add a 3 on the left side of the equation with \(\mathrm{O_2}\).
\(\mathrm{C_2 H_5 OH + 3 O_2 = 2 C O_2 + 3 H_2 O}\)
How to balance an equation with Linear Systems?
Complex equations can be balanced by an algebraic approach based on solving a series of linear equations as if there are multiple compounds in the equation. Let's balance this equation with a linear system step by step.
\(\mathrm{K_4 Fe (CN)_6 + H_2 SO_4 + H_2 O = K_2SO_4 + Fe SO_4 + (NH_4)_2SO_4 + CO}\)
First of all, assign each coefficient with the variables. Coefficients represent the mole ratios and basic unit in the equation that are balanced:
\(\mathrm{a K_4 Fe (CN)_6 + b H_2 SO_4 + c H_2 O = d K_2 SO_4 + e Fe SO_4 + f (NH_4) 2 SO_4 + g CO}\)
On each side of the equation, there must be the same amount of each atom. Count atoms for each element and make all sides equal.
 K: 4 a = 2 d
 Fe: 1 a = 1 e
 C: a = 6 g
 N: a = 3 f
 H: 2 b + 2 c = 8 f
 S: b = d + e + f
 O: 4 b + c = 4 d + 4 e + 4 f + g
Substitute directly to solve the system.
 d = 2 a
 e = a
 g = 6 a
 f = 3 a
 b = 6 a
 c = 6 a
All coefficients depend on parameter a, so only choose a=1, which gives a number that allows all small whole numbers.
a = 1, b = 6, c = 6, d = 2, e = 1, f = 3, g = 6
In the end, you will get the balanced equation:
\(\mathrm{K_4 Fe (CN)_6 + 6 H_2 SO_4 + 6 H_2 O = 2 K_2 SO_4 + Fe SO_4 + 3 (NH_4)_2 SO_4 + 6 CO}\)
You can use both methods to achieve a more realistic algorithm to accelerate the process.
Identify elements that appear in each compound. Begin with the one that has a greater index, which allows you to continue working with integer and add a variable.
\(\mathrm{a K_4 Fe (CN)_6 + H_2 SO_4 + H_2 S = K_2 SO_4 + Fe SO_4 + (NH_4)_2 SO_4 + CO}\)
Fe SO 4 must be 1 a due to Fe, K 2 SO 4 must be 2 a due to K, (NH 4) 2 SO 4 must be 3 a due N, and CO must be 6 a due to C. It excludes the first four system equations. Whatever the coefficients are, it is already understood that these proportions must carry:
\(\mathrm{a K_4 Fe (CN)_6 + H_2 SO_4 + H_2 O = 2a K_2 SO_4 + a Fe SO_4 + 3a (NH_4)_2 SO_4 + 6a CO}\)
You can now write the equations and solve them more easily. For this specific equation, you may remember that the addition of Sulphur is 6a for H 2 SO 4 and, ultimately, with the inclusion of hydrogen, the remaining 6a for H 2 SO 4 can be found.
\(\mathrm{K_4 Fe (CN)_6 + 6 H_2 SO_4 + 6 H_2 O = 2 K_2 SO_4 + Fe SO_4 + 3 (NH_4)_2 SO_4 + 6 CO}\)

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