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 BaCO3 + 1 Pb3O4 + 1 SrCO3 + 1 O2 = 1 Ba1000Pb875Sr125O3000 + 1 CO2
For each element, we check if the number of atoms is balanced on both sides of the equation. Ba is not balanced: 1 atom in reagents and 1000 atoms in products. In order to balance Ba on both sides we: Multiply coefficient for BaCO3 by 1000 1000 BaCO3 + 1 Pb3O4 + 1 SrCO3 + 1 O2 = 1 Ba1000Pb875Sr125O3000 + 1 CO2
Pb is not balanced: 3 atoms in reagents and 875 atoms in products. In order to balance Pb on both sides we: Multiply coefficient for Pb3O4 by 875 Multiply coefficient for Ba1000Pb875Sr125O3000 by 3 1000 BaCO3 + 875 Pb3O4 + 1 SrCO3 + 1 O2 = 3 Ba1000Pb875Sr125O3000 + 1 CO2
Sr is not balanced: 1 atom in reagents and 375 atoms in products. In order to balance Sr on both sides we: Multiply coefficient for SrCO3 by 375 1000 BaCO3 + 875 Pb3O4 + 375 SrCO3 + 1 O2 = 3 Ba1000Pb875Sr125O3000 + 1 CO2
Ba is not balanced: 1000 atoms in reagents and 3000 atoms in products. In order to balance Ba on both sides we: Multiply coefficient for BaCO3 by 3 3000 BaCO3 + 875 Pb3O4 + 375 SrCO3 + 1 O2 = 3 Ba1000Pb875Sr125O3000 + 1 CO2
C is not balanced: 3375 atoms in reagents and 1 atom in products. In order to balance C on both sides we: Multiply coefficient for CO2 by 3375 3000 BaCO3 + 875 Pb3O4 + 375 SrCO3 + 1 O2 = 3 Ba1000Pb875Sr125O3000 + 3375 CO2
O is not balanced: 13627 atoms in reagents and 15750 atoms in products. In order to balance O on both sides we: Multiply coefficient for O2 by 2125 Multiply coefficient(s) for Ba1000Pb875Sr125O3000, CO2, BaCO3, Pb3O4, SrCO3 by 2 6000 BaCO3 + 1750 Pb3O4 + 750 SrCO3 + 2125 O2 = 6 Ba1000Pb875Sr125O3000 + 6750 CO2
Ba is balanced: 6000 atoms in reagents and 6000 atoms in products. Pb is balanced: 5250 atoms in reagents and 5250 atoms in products. Sr is balanced: 750 atoms in reagents and 750 atoms in products. C is balanced: 6750 atoms in reagents and 6750 atoms in products. All atoms are now balanced and the whole equation is fully balanced: 6000 BaCO3 + 1750 Pb3O4 + 750 SrCO3 + 2125 O2 = 6 Ba1000Pb875Sr125O3000 + 6750 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 BaCO3 + b Pb3O4 + c SrCO3 + d O2 = e Ba1000Pb875Sr125O3000 + f CO2
Now we write down algebraic equations to balance of each atom: Ba: a * 1 = e * 1000 C: a * 1 + c * 1 = f * 1 O: a * 3 + b * 4 + c * 3 + d * 2 = e * 3000 + f * 2 Pb: b * 3 = e * 875 Sr: c * 1 = e * 125
Now we assign a=1 and solve the system of linear algebra equations: a = e000 a + c = f a * 3 + b * 4 + c * 3 + d * 2 = e * 3000 + f * 2 b * 3 = e * 875 c = e25 a = 1
Solving this linear algebra system we arrive at: a = 1 b = 0.29166666666667 c = 0.125 d = 0.35416666666667 e = 0.001 f = 1.125
To get to integer coefficients we multiply all variable by 6000 a = 6000 b = 1750 c = 750 d = 2125 e = 6 f = 6750
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: 6000 BaCO3 + 1750 Pb3O4 + 750 SrCO3 + 2125 O2 = 6 Ba1000Pb875Sr125O3000 + 6750 CO2
Direct link to this balanced equation:
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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.
