Balancing step by step using the inspection method
Let's balance this equation using the inspection method. First, we set all coefficients to 1: 1 K2CO3 + 1 Na2CO3 + 1 Nb2O5 + 1 Sb2O3 + 1 TiO2 + 1 Bi2O3 + 1 O = 1 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
For each element, we check if the number of atoms is balanced on both sides of the equation. K is not balanced: 2 atoms in reagents and 4761 atoms in products. In order to balance K on both sides we: Multiply coefficient for K2CO3 by 4761 Multiply coefficient for K4761Na5189Nb9405Sb495O30000Bi50Ti100 by 2 4761 K2CO3 + 1 Na2CO3 + 1 Nb2O5 + 1 Sb2O3 + 1 TiO2 + 1 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
Na is not balanced: 2 atoms in reagents and 10378 atoms in products. In order to balance Na on both sides we: Multiply coefficient for Na2CO3 by 5189 4761 K2CO3 + 5189 Na2CO3 + 1 Nb2O5 + 1 Sb2O3 + 1 TiO2 + 1 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
Nb is not balanced: 2 atoms in reagents and 18810 atoms in products. In order to balance Nb on both sides we: Multiply coefficient for Nb2O5 by 9405 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 1 Sb2O3 + 1 TiO2 + 1 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
Sb is not balanced: 2 atoms in reagents and 990 atoms in products. In order to balance Sb on both sides we: Multiply coefficient for Sb2O3 by 495 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 1 TiO2 + 1 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
Ti is not balanced: 1 atom in reagents and 200 atoms in products. In order to balance Ti on both sides we: Multiply coefficient for TiO2 by 200 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 200 TiO2 + 1 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
Bi is not balanced: 2 atoms in reagents and 100 atoms in products. In order to balance Bi on both sides we: Multiply coefficient for Bi2O3 by 50 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 200 TiO2 + 50 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 1 CO2
C is not balanced: 9950 atoms in reagents and 1 atom in products. In order to balance C on both sides we: Multiply coefficient for CO2 by 9950 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 200 TiO2 + 50 Bi2O3 + 1 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 9950 CO2
O is not balanced: 78911 atoms in reagents and 79900 atoms in products. In order to balance O on both sides we: Multiply coefficient for O by 990 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 200 TiO2 + 50 Bi2O3 + 990 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 9950 CO2
All atoms are now balanced and the whole equation is fully balanced: 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 200 TiO2 + 50 Bi2O3 + 990 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 9950 CO2
Balancing step by step using the algebraic method
Let's balance this equation using the algebraic method. First, we set all coefficients to variables a, b, c, d, ... a K2CO3 + b Na2CO3 + c Nb2O5 + d Sb2O3 + e TiO2 + f Bi2O3 + g O = h K4761Na5189Nb9405Sb495O30000Bi50Ti100 + i CO2
Now we write down algebraic equations to balance of each atom: K: a * 2 = h * 4761 C: a * 1 + b * 1 = i * 1 O: a * 3 + b * 3 + c * 5 + d * 3 + e * 2 + f * 3 + g * 1 = h * 30000 + i * 2 Na: b * 2 = h * 5189 Nb: c * 2 = h * 9405 Sb: d * 2 = h * 495 Ti: e * 1 = h * 100 Bi: f * 2 = h * 50
Now we assign a=1 and solve the system of linear algebra equations: a * 2 = h * 4761 a + b = i a * 3 + b * 3 + c * 5 + d * 3 + e * 2 + f * 3 + g = h * 30000 + i * 2 b * 2 = h * 5189 c * 2 = h * 9405 d * 2 = h * 495 e = h00 f * 2 = h * 50 a = 1
Solving this linear algebra system we arrive at: a = 1 b = 1.0898970804453 c = 1.9754253308129 d = 0.10396975425331 e = 0.042007981516488 f = 0.010501995379122 g = 0.20793950850662 h = 0.00042007981516488 i = 2.0898970804453
To get to integer coefficients we multiply all variable by 4761 a = 4761 b = 5189 c = 9405 d = 495 e = 200 f = 50 g = 990 h = 2 i = 9950
Now we substitute the variables in the original equations with the values obtained by solving the linear algebra system and arrive at the fully balanced equation: 4761 K2CO3 + 5189 Na2CO3 + 9405 Nb2O5 + 495 Sb2O3 + 200 TiO2 + 50 Bi2O3 + 990 O = 2 K4761Na5189Nb9405Sb495O30000Bi50Ti100 + 9950 CO2
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Instructions on balancing chemical equations:
Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
To enter an electron into a chemical equation use {-} or e
To enter an ion, specify charge after the compound in curly brackets: {+3} or {3+} or {3}. Example: Fe{3+} + I{-} = Fe{2+} + I2
Substitute immutable groups in chemical compounds to avoid ambiguity. For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will
Compound states [like (s) (aq) or (g)] are not required.
If you do not know what products are, enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.
Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. The limiting reagent row will be highlighted in pink.
Examples of complete chemical equations to balance:
A chemical equation represents a chemical reaction. It shows the reactants (substances that start a reaction) and products (substances formed by the reaction). For example, in the reaction of hydrogen (H₂) with oxygen (O₂) to form water (H₂O), the chemical equation is:
However, this equation isn't balanced because the number of atoms for each element is not the same on both sides of the equation. A balanced equation obeys the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.
Balancing with inspection or trial and error method
This is the most straightforward method. It involves looking at the equation and adjusting the coefficients to get the same number of each type of atom on both sides of the equation.
Best for: Simple equations with a small number of atoms.
Process: Start with the most complex molecule or the one with the most elements, and adjust the coefficients of the reactants and products until the equation is balanced.
Count the number of H and O atoms on both sides. There are 2 H atoms on the left and 2 H atom on the right. There are 2 O atoms on the left and 1 O atom on the right.
Balance the oxygen atoms by placing a coefficient of 2 in front of H2O:
Check the balance. Now, both sides have 4 H atoms and 2 O atoms. The equation is balanced.
Balancing with algebraic method
This method uses algebraic equations to find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, y, z), and a series of equations are set up based on the number of each type of atom.
Best for: Equations that are more complex and not easily balanced by inspection.
Process: Assign variables to each coefficient, write equations for each element, and then solve the system of equations to find the values of the variables.
Assign one of the coefficients to 1 and solve the system.
a = 1
c = 2 a = 2
d = 6 a / 2 = 4
b = (2 c + d) / 2 = (2 * 2 + 3) / 2 = 3.5
Adjust coefficient to make sure all of them are integers. b = 3.5 so we need to multiple all coefficient by 2 to arrive at the balanced equation with integer coefficients:
Useful for redox reactions, this method involves balancing the equation based on the change in oxidation numbers.
Best For: Redox reactions where electron transfer occurs.
Process: identify the oxidation numbers, determine the changes in oxidation state, balance the atoms that change their oxidation state, and then balance the remaining atoms and charges.
This method separates the reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced separately and then combined.
Best for: complex redox reactions, especially in acidic or basic solutions.
Process: split the reaction into two half-reactions, balance the atoms and charges in each half-reaction, and then combine the half-reactions, ensuring that electrons are balanced.
