How to balance the reaction equation of potassium permanganate, calcium oxalate, and sulfuric acid step by step? I have been desperately trying to balance the following equation, and finally (ultima ratio) used an online program to get it done (posted the same question there as well).
$$\ce{KMnO4 + CaC2O4 + H2SO4 -&gt; MnSO4 + K2SO4 + CaSO4 + CO2 + H2O}\tag{I}$$

No worries, balancing the K, Mn, Ca, C and S - but by then the H and O got out of my control. So this is how far I got myself:
$$\begin{multline}\ce{2KMnO4 + CaC2O4 + 4H2SO4 -&gt;\\
 2MnSO4 + K2SO4 + CaSO4 + 2CO2 + 4H2O}\tag{II}\end{multline}$$

So the only thing left to be balanced, was the O, having 28 on the reagents' side and 24 on the product side. I started off, getting up to 36:33 O (reagents:products), and I felt that I was getting nowhere.

Anyone who would guide me through the steps following the balancing until equation II above?
This is a redox reaction. While it can be balanced by the method you describe, it is much more reliable to balance it by the half-reaction method.

Step 1 - Identify your spectator ions (so we do not spend time trying to balance them)

Three ions appear unchanged on both sides of the reaction: $\ce{K+},\ \ce{Ca^2+},\ \&amp; \ \ce{SO4^2-}$

Step 2 - Identify your redox pairs.

You have one manganese species on each side. They must constitute one half-reaction:
$$\ce{MnO4- -&gt; Mn^2+}$$
Likewise, you have on carbon species on each side. They must constitute the other half reaction:
$$\ce{C2O4^2- -&gt; CO2}$$
For each half reaction:

Step 3 - Balance O by adding $\ce{H2O}$
Step 4 - Balance H by adding $\ce{H+}$
Step 5 - Balance charge by adding $\ce{e-}$
Step 6 - Combine the half reactions

Can you take it from here?