![]() Our chemical equation balancer calculator uses the Gauss-Jordan elimination algorithm for solving a system of linear equations. However, in cases where we are dealing with more complex reactions involving many compounds, it is preferable to balance the equations using more advanced algebraic methods based on solving a system of linear equations. Thus, we have the following coefficients: a=1, b=4, c=1, d=5Īs the result we have the following balanced equation: Let’s solve the system by direct substitution and find the values of the remaining variables: To quickly solve this system of equations, we assign a numerical value to one of the variables. It is necessary to find such a solution that all the coefficients have the form of the smallest possible integers. This system has several solutions, since there are more variables than equations. It remains for us to solve the above system of linear equations in order to find the numerical values of the coefficients. As can be easily checked by substituting in the previous equations, this equality is a consequence of them, not an independent relation. Since the left and right sides must contain the same number of hydrogen atoms, we get: 2 b =3 c+ d. For example, on the left we have 2 b hydrogen atoms (2 in each H2O molecule), while on the right we have 3 c+ d hydrogen atoms (3 in each H3PO4 molecule and 1 in each HF molecule). Next, we equate the number of atoms of each element in the left and right sides of the equation. Put a factor in front of the single carbon atom on the right side of the equation to balance it with the 2 carbons on the left side of the equation: So the first thing to do in our case is to balance the carbon. If it is necessary to balance several elements, we first choose one that is part of only one molecule of reactants and one molecule of reaction products. Note that usually hydrogen and oxygen are part of several molecules at once, so it is better to balance them last. ![]() On the right side we have 1 carbon atom, 2 hydrogen atoms and 3 oxygen atoms: C=1, H=2, O=3.On the left side we have 2 carbon atoms, 6 hydrogen atoms and 2 oxygen atoms: C=2, H=6, O=2.Consider the subscripts next to each element to determine the total number of atoms. Let us first write down the number of atoms of each element for both sides of the equation. Relatively simple equations like this one can be balanced by inspection and directly fitting the stoichiometric coefficients. Enter the value 1 in the bottom right cell I8, and fill column K with zeros and 1 as shown.As we can see, atoms of only three chemical elements are involved in this reaction. For the products, column G chromium pentoxide has -5 O and -1 Cr, column H potassium sulphate has -2 K, -1 S and -4 O, and column I water has -2 H and -1 O. Thus in column D, hydrogen peroxide has 2 H and 2 O atoms in column E potassium dichromate has 2 K, 7 O and 2 Cr and in column F sulphuric acid has 2 H, 4 O and 1 S. The table is filled with the numbers of each type of atom in the reactant and product compounds, reactants being positive and products negative. In column J are entered each of the atomic species in the reaction, ie H,K,O,Cr & S ( their order doesn't matter ) and x at the end. Next is a table in block D3:K8 which will contain the input matrix, with the coefficient headings a-f at the top of the block. Column C is for the calculated results and can be left empty to begin with. The reactants and products are entered in column A and their coefficients a,b,c, d,e, & f in column B. ![]() The next step is to set up a table in the spreadsheet as shown in Table 1. a H2O2 + b K2Cr2O7 + c H2SO4 → d CrO5 + e K2SO4 + f H2O ![]() To begin with, we write out the reactant and product species formulae and apply algebraic coefficients to each: Reactants ProductsĮg.
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