To balance equation, it is important to understand the fundamentals. Introduce to them the periodic table and teach them that the period indicates the number of electronic shells and group represents the number of valance electrons.
Next they need to know the charges of each elements eg. Cu (1+ or 2+) Ag (1+) SO4 (2-) from them once they get the formula of the compound correctly then they will be able to start balancing the equation to balance, write make sure the formula of each compound is correctly written
then write down both the left and right hand side of each of the number of element.
Finally, find the lowest common multiple of each element on each side takes abit of practice but the important this is to make sure that the formula of each compound is correct.
Ensure that the chemical formula of the reactants and products are correct
You can only add numbers in front of the chemical formula
Balance polyatomic ions as a whole if it remains in tact as a reactant and product. For example, HNO3 + Mg –> Mg(NO3)2 + H2 Balance NO3- as a whole (no need to split into N and O). So we have 1 NO3- in reactant, but 2 in product. so put a 2 in front of HNO3. That nicely balance H2.
Balance the element which is found in more than one reactant/product last. Typically, but not always oxygen/hydrogen. For example, in the equation Pb(NO3)2 –> PbO + NO2 + O2 Oxygen is found in all the three products, so balance this last. We cannot balance NO3- as a whole (See point 3) as it did not remain as NO3- in product but is broken up. To balance this equation, start with Pb (which is balanced). There are 2 N in reactant, so put a 2 in front of NO2. So now we have Pb(NO3)2 –> PbO + 2NO2 + O2. This gives us 6 oxygen in reactant, but 7 oxygen (1 from Pb), 4 from NO2, and 1 from O2) in products. So we need to increase the number of oxygen in reactant and we do this by putting a 2 in front of Pb(NO3)2. Rebalance by putting a 2 in front of PbO, 4 in front of NO2 and now the number of oxygen atoms nicely balanced out.
Special case: Combustion of organic compounds C3H6 + O2 –> CO2 + H2O Using point 4, balance oxygen last. So we have C3H6 + O2 –> 3CO2 + 3H2O Total oxygen in products is 9. We need a number in front of reactant O2 to give 9 oxygen, so we put 9/2 (4.5). As we usually balance using whole numbers, we multiply by 2 throughout to get 2C3H6 + 9O2 –> 6CO2 + 6H2O