![]() ![]() Let's count - on the left side of equation we have 9 atoms of oxygen in calcium carbonate. We can omit oxygen atoms in PO 4 group as they are already balanced. There are 6 hydrogens on the left - once again we need to add 3 on the right side of the equation, this time in front of H 2O:ģCaCO 3 + 2H 3 PO 4 → 1Ca 3( PO 4) 2 + 3H 2O + 3CO 2 There are three carbon atoms on the left, so we add 3 in front of CO 2 on the right:ģCaCO 3 + 2H 3 PO 4 → 1Ca 3( PO 4) 2 + H 2O + 3CO 2 What is left to balance? Carbon, oxygen and hydrogen. We also need three calcium atoms on the left:ģCaCO 3 + 2H 3 PO 4 → 1Ca 3( PO 4) 2 + H 2O + CO 2 Now, instead of calculating individual atoms of phosphorus and oxygen we can balance PO 4 as a whole - there are two such groups in calcium phosphate, thus we need two phosphoric acid molecules on the left side of the equation:ĬaCO 3 + 2H 3 PO 4 → 1Ca 3( PO 4) 2 + H 2O + CO 2 We will treat PO 4 group (note that to be correct we should treat it as a ion, which we will neglect at the moment in notation) as one large entity (we will make it bold so that it will be easier to spot):ĬaCO 3 + H 3 PO 4 → 1Ca 3( PO 4) 2 + H 2O + CO 2 Obviously the most complicated molecule - one that we will start with - is that of calcium phosphate, so let's assume there is one such molecule in the balanced equation:ĬaCO 3 + H 3PO 4 → 1Ca 3(PO 4) 2 + H 2O + CO 2 We will use it to balance equation of reaction between calcium carbonate and phosphoric acid:ĬaCO 3 + H 3PO 4 → Ca 3(PO 4) 2 + H 2O + CO 2 Treat polyatomic groups that don't change in the reaction as if they were kind of large 'atoms' - instead of balancing atoms in the group individually, balance group as a whole. ![]() ![]() To do so let's multiply all coefficients by two: Right now there are same numbers of atoms on both sides of the equation - but to finish balancing we have to remove the fraction. Remember, we can use fractions at this stage: Two are already present in the ethylene glycol, so we need five more - 2 1/2 oxygen molecule will do. To balance equation thus we need seven oxygen atoms on the left. There are seven atoms of oxygen on the right - four in molecules of carbon dioxide and three in water molecules. There are two atoms of carbon in ethylene glycol molecule, so let's add 2 in front of CO2.Ĭarbon is balanced, what about hydrogen? There are six hydrogen atoms on the left side of the reaction, so we need three water molecules on the right to balance hydrogen: Let's start balancing with carbon and hydrogen. Technically speaking if there is no coefficient, it means there is only one molecule of the compound, but we will put ones into the equations now to differentiate between molecules we have already tried to balance, and those that are not touched yet.īoth carbon and hydrogen are present in two molecules, oxygen is present in all four molecules. The most complicated molecule is that of ethylene glycol, so let's put 1 in front of it. Let's try with ethylene glycol combustion: Once all atoms and charges are balanced, remove fractions and find set of the smallest coefficients.Balance atoms present in most compounds at the end.Don't be afraid of using fractions at this stage. Start balancing atoms present in the lowest number of compounds.Assume there is only one molecule of this compound taking part in the reaction. Start with the most complicated compound.How to balance an equation by inspection? You should follow these four basic rules: More complicated ones require experience - if you don't have it and you are not afraid of relatively simple mathematics, you will probably find the algebraic method more effective. Balancing chemical reaction equations by inspectionīalancing by inspection is the most basic method used.
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