Question

How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.

Answer 1

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