step by step, show how to balance C2H5OH + K2Cr2O7 + H2SO4 = Cr2(SO4)3 + K2SO4 + CH3COOH + H2O

Answer 1

Since can't draw diagrams on this forum here is the best I can do. Below are the electrons changes; you can draw the lines/arrows and whatever else you need to do but this is the heart of the problem.
C2H5OH. Both carbons are -4 total.
CH3COOH. Both carbons are zero each(and total)

K2Cr2O7. Both Cr atoms are +12 total.
Cr2(SO4)3. Both Cr atoms are +6 total

Both carbons change from -4 to zero or loss of 4 electrons.
Both Chromiums change from +12 to +6 or gain of 6 electrons.
Therefore, you want to multiply C half rxn by 3 and multiply Cr half rxn by 2 (LCD is 12). So it ends up like this for the major part of it.

3C2H5OH + 2K2Cr2O7 + H2SO4 = 2Cr2(SO4)3 + K2SO4 + 3CH3COOH + H2O

That's the hard part. Balance the remainder by inspection.
Some hints. The only place I see K on the left is K2Cr2O7 and since that 2 is in concrete, that makes K2SO4 simply 2K2SO4.. Now look on the right. The SO4 for 2Cr2(SO4)3 is in concrete and the 2K2SO4 is now in concrete (after adding the 2), so you know you need that many SO4 on the left. And look, the H2SO4 is the ONLY place I see SO4 on the left. Yay. So plug that number in. That makes H and O the only thing left. Note that I didn't count the O in SO4 since I balanced sulfate by itself. Let me know how you far. I'll be glad to check you answer.