Answer:
3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O
Explanation:
To balance, start of with the groups that are common on both sides of the reaction equation;
In this case, these are the SO4 groups treating them as single units;
There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;
Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;
Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;
Lastly, we can deal with the O and H;
Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;
7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;
If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;
In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;
Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;
Thus we end up with:
3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O
Rules of thumb:
- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units
- Start of with balancing these common groups
- Thereafter, balance the atoms that appear in only one reactant and one product
- Proceed to balance the atoms that appear in more than one reactant or product
- Typically, you should deal with the O and H last
