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 C9 + 1 H8 + 1 O7 + 1 N6 + 1 He5 + 1 K4 = 1 C2H3O4N5He6K7
For each element, we check if the number of atoms is balanced on both sides of the equation. C is not balanced: 9 atoms in reagents and 2 atoms in products. In order to balance C on both sides we: Multiply coefficient for C9 by 2 Multiply coefficient for C2H3O4N5He6K7 by 9 2 C9 + 1 H8 + 1 O7 + 1 N6 + 1 He5 + 1 K4 = 9 C2H3O4N5He6K7
H is not balanced: 8 atoms in reagents and 27 atoms in products. In order to balance H on both sides we: Multiply coefficient for H8 by 27 Multiply coefficient for C2H3O4N5He6K7 by 8 2 C9 + 27 H8 + 1 O7 + 1 N6 + 1 He5 + 1 K4 = 72 C2H3O4N5He6K7
O is not balanced: 7 atoms in reagents and 288 atoms in products. In order to balance O on both sides we: Multiply coefficient for O7 by 288 Multiply coefficient for C2H3O4N5He6K7 by 7 2 C9 + 27 H8 + 288 O7 + 1 N6 + 1 He5 + 1 K4 = 504 C2H3O4N5He6K7
N is not balanced: 6 atoms in reagents and 2520 atoms in products. In order to balance N on both sides we: Multiply coefficient for N6 by 420 2 C9 + 27 H8 + 288 O7 + 420 N6 + 1 He5 + 1 K4 = 504 C2H3O4N5He6K7
He is not balanced: 5 atoms in reagents and 3024 atoms in products. In order to balance He on both sides we: Multiply coefficient for He5 by 3024 Multiply coefficient for C2H3O4N5He6K7 by 5 2 C9 + 27 H8 + 288 O7 + 420 N6 + 3024 He5 + 1 K4 = 2520 C2H3O4N5He6K7
K is not balanced: 4 atoms in reagents and 17640 atoms in products. In order to balance K on both sides we: Multiply coefficient for K4 by 4410 2 C9 + 27 H8 + 288 O7 + 420 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
C is not balanced: 18 atoms in reagents and 5040 atoms in products. In order to balance C on both sides we: Multiply coefficient for C9 by 280 560 C9 + 27 H8 + 288 O7 + 420 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
H is not balanced: 216 atoms in reagents and 7560 atoms in products. In order to balance H on both sides we: Multiply coefficient for H8 by 35 560 C9 + 945 H8 + 288 O7 + 420 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
O is not balanced: 2016 atoms in reagents and 10080 atoms in products. In order to balance O on both sides we: Multiply coefficient for O7 by 5 560 C9 + 945 H8 + 1440 O7 + 420 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
N is not balanced: 2520 atoms in reagents and 12600 atoms in products. In order to balance N on both sides we: Multiply coefficient for N6 by 5 560 C9 + 945 H8 + 1440 O7 + 2100 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
All atoms are now balanced and the whole equation is fully balanced: 560 C9 + 945 H8 + 1440 O7 + 2100 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
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 C9 + b H8 + c O7 + d N6 + e He5 + f K4 = g C2H3O4N5He6K7
Now we write down algebraic equations to balance of each atom: C: a * 9 = g * 2 H: b * 8 = g * 3 O: c * 7 = g * 4 N: d * 6 = g * 5 He: e * 5 = g * 6 K: f * 4 = g * 7
Now we assign a=1 and solve the system of linear algebra equations: a * 9 = g * 2 b * 8 = g * 3 c * 7 = g * 4 d * 6 = g * 5 e * 5 = g * 6 f * 4 = g * 7 a = 1
Solving this linear algebra system we arrive at: a = 1 b = 1.6875 c = 2.5714285714286 d = 3.75 e = 5.4 f = 7.875 g = 4.5
To get to integer coefficients we multiply all variable by 560 a = 560 b = 945 c = 1440 d = 2100 e = 3024 f = 4410 g = 2520
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: 560 C9 + 945 H8 + 1440 O7 + 2100 N6 + 3024 He5 + 4410 K4 = 2520 C2H3O4N5He6K7
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Compound may be misspelled: C9 -> C Compound may be misspelled: H8 -> H Compound may be misspelled: O7 -> O Compound may be misspelled: N6 -> N2 Compound may be misspelled: K4 -> K
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.
