One of the most difficult part of chemistry is chemical equations, how to correctly write a balanced chemical equation. Obviously without a correctly written chemical equation, any calculation in chemistry is an error. In other words, chemistry is equation and equation is chemistry. A balanced equation tells the chemist or the student what is happening in the reaction. But students and even some teachers sometimes fumble at writing a balanced equation possibly because the only way they knew of is the usual trial and error method! I thought I’m alone until I met two lecturers who gave wrong chemical equations to their students!

Without any further ado, I introduce to you MY fascinating, cre8ive, and easy way to balance any chemical equation!

Please Note: remember that when balancing an equation:

The formulae of the reactants and products in an equation are fixed and cannot be altered. In other words, don’t change the subscripts.
Only write the appropriate numbers of moles of the reactants and products concerned. In other words, you can rightly change the coefficients of the reactants and products.
My Methods of Balancing Equations Without Radicals

Only three simple steps and you are there:
Step 1. Pick out the element with the highest subscript on either side of the equation (reactant/LHS or product/RHS)
Step 2. Multiply this subscript with the lowest subscript of the same element on the opposite side of the equation.
Step 3. Place the answer in front (as the number of moles or coefficient) of the element or compound with the latter subscript…
and that is all, finish it up by doing an “atom count”!

For instance, 1: the combustion of ammonia gas in air

NH3(g) + O2(g) —> NO(g) + H2O(g) . To balance this equation, let’s explicitly follow the three steps given above.

1. The element Hydrogen on the LHS has the highest subscript (3).

2. This 3 multiply by 2 (the subscript of the same Hydrogen on the RHS). i.e 3 × 2 = 6

3. Write this 6 in front of, or as a number of moles of H2O because that is where we took the last subscript, 2 from: 6H2O. What you have done:

NH3 + O2 —> NO + 6H2O

No doubt it’s very easy to finish up from here. By atom count, we have 12 hydrogen on RHS but only 3 on the LHS which with simple math, we need 4 in front of NH3 (4NH3). This changes the number of nitrogen and ultimately the number of Oxygen atoms. Eventually we arrive at the correct equation:

4NH3(g) + 5O2(g) + 4NO(g) + 6H2O(l)

e.g 2. The reaction between hydrochloric acid and calcium carbonate

HCl(aq) + CaCO3(s) —> CaCl(aq) + H2O(l) + CO2(g)
By mere looking, I know Oxygen is balanced (at least for now) then I move on to Hydrogen. On multiplying the 2 on the RHS with 1 on the LHS I arrived at

2HCl(l) + CaCO3(s) —> CaCl2(aq) + H2O(l) + CO2(g)

And this is the balanced equation! Yeah, it’s that simple!

e.g 3. KMnO4(aq) + HCl(aq) —> MnCl2(aq) KCl(aq) + H2O(l) + Cl2(g)

Multiplying 4×1 and writing the answer 4 as the number of moles of water requires writing 8 as the number of moles of HCl which triggers a chain of imbalances, though very very easy to deal with. Eventually you will arrive at the correct equation:

2KMnO4(aq) + 16HCl(l) —> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g).

For the avoidance of doubt, please pay close attention as I explain below.

Upon writing 8 as the number of moles of HCl, Cl becomes 8 on the LHS as against 5 on the RHS. Now, by mere reasoning, the only way to balance the Cl atoms at this point is to write 2 in front of MnCl2 and 2 in front of KCl resulting in 2MnCl2 and 2KCl respectively. Without any question, you know that you have to place 2 in front of KMnO4, thus the 2KMnO4. Here is what you have done thus far:

2KMnO4(aq) + 8HCl(l) —> 2MnCl2(aq) + 2KCl + 4H2O + Cl2

Can you see the chain of imbalances? But very simple to solve!

So, to have 8 atoms of Oxygen on the RHS, I (or you) need to change the 4 moles of water (4H2O) to 8 moles ( 8H2 O), resulting to 16 atoms of H2. Finally you change the 8 moles of HCl to 16 moles, which tells you to place 5 in front of chlorine thus giving you:

2KMnO4(aq) + 16HCl(l) —> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g).

The above is one of the few complex equations without radicals you will come across. However, with practice you will become more adept at balancing equations faster with this method.

Now the ball is in your court. Compare your newly learned method with the one(s) you knew to solve the above equation and the exercises below?

EXERCISES

1. C2H2(g) + O2(g) —> CO2(g) + O2(g)

2. Zn(s) + HCl(aq) —> ZnCl2(aq) + H2(g)

3. NaNO3(s) –heat–> NaNO2(s) + O2(g)

4.Pb(NO3)2(aq) + HCl(aq) —> PbCl2(s) + HNO3(aq)

5. KClO3(s) —> KClO2(s) + O2(g)

6. Fe(s) + H2O(g) —> Fe3O4(s) + H2(g)

#sometakeaways

In any equation with Oxygen, start balancing with Oxygen atoms first.

If in any equation one element has the same number of subscript on one side-as in exercise 5- consider using one of these subscripts as a start, by writing this number as the number of moles where appropriate, before using this method.

Now if nothing else, I’m sure I have given you some clues on how and where to start with when it comes to balancing an equation.

For equations involving radicals, I will give you some tips in my next article, balancing equations made easy part II. Thanks.

Source: https://eneyellowacademy.com/2017/01/18/chemical-equations-balancing-equations-made-easy-part-i/

And here is the part 2
https://eneyellowacademy.com/2017/01/22/chemical-equations-balancing-equations-made-easy-part-ii/