The chemical method of analysis in determination of the blood alcohol content %BAC is:

K2Cr2O7 + H2SO4 + C2H5OH ---> Cr2(SO4)3 + K2SO4 + CH3COOH + H2O

What are the stoichiometric coefficients for the reaction above?

Cr is reduced from +6 each on the left to +3 each on the right.
C (in C2H5OH) is oxidized from -2 each on the left to zero for each on the right. Draw a line from Cr on the left to Cr on the right. On the left write +12 above the 2Cr (that is +6 each x 2 atoms = +12) and on the right write +6 above the Cr2 (that is +3 each x 2 = +6).
Next, draw a line from C on the left to C on the right. On the left write -4 above the C2 (that is -2 each x 2 = 4) and on the right write 0 (that is 0 each x 2 = 0).
Here is how you finnish.
1. Determine the loss and gain of electrons.
2. Multiply the loss and gain by coefficients that will make the loss equal to the gain. I found 3 and 4 to be the numbers used to multiply to make 12 electrons lost and 12 electrons gained.
3. Multiply those compounds connected with the lines you have drawn by the coefficients used to make the e gain = e lost; i.e., you will multiply one set by 3 and one set by 4.
4. By inspection, balance those atoms not changed by oxidation or reductin.
Post your work if you get stuck. I shall be happy to check your finished equation if you show the intermediate steps that lead to the finished product. I hope this helps.

Thanks for your help. I will post my work if I need more help.

coefficients are 2+8+3 --> 2+2+3+11

To determine the stoichiometric coefficients for the given chemical reaction, you need to balance the equation by ensuring that the number of atoms on both sides of the reaction is the same. Here's a step-by-step guide:

1. Start by balancing the elements that appear in only one compound on each side of the equation. In this case, we have Cr, C, H, O, K, and S.

2. Balance the Cr atoms: On the left side, we have 2 Cr atoms, and on the right side, we have Cr2. To balance Cr, put a coefficient of 2 in front of Cr2(SO4)3 on the right side. This gives us:

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

3. Balance the C atoms: On the left side, we have 2 C atoms in C2H5OH, and on the right side, we have 1 C atom in CH3COOH. To balance C, put a coefficient of 2 in front of CH3COOH on the right side. This gives us:

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

4. Balance the H atoms: On the left side, we have 6 H atoms in C2H5OH and 2 H atoms in H2O, adding up to a total of 8 H atoms. On the right side, we have 4 H atoms in 2CH3COOH and 2 H atoms in H2O, adding up to a total of 6 H atoms. To balance H, put a coefficient of 4 in front of H2O on the left side. This gives us:

K2Cr2O7 + H2SO4 + C2H5OH ---> 2Cr2(SO4)3 + K2SO4 + 2CH3COOH + 4H2O

5. Balance the O atoms: On the left side, we have 7 O atoms in K2Cr2O7 and 4 O atoms in H2SO4, adding up to a total of 11 O atoms. On the right side, we have 24 O atoms in 2Cr2(SO4)3, 4 O atoms in K2SO4, 4 O atoms in 2CH3COOH, and 8 O atoms in 4H2O, adding up to a total of 40 O atoms. To balance O, put a coefficient of 12 in front of K2Cr2O7 on the left side. This gives us the final balanced equation:

12K2Cr2O7 + 3H2SO4 + 8C2H5OH ---> 24Cr2(SO4)3 + 4K2SO4 + 4CH3COOH + 8H2O

Therefore, the stoichiometric coefficients for the reaction are:

K2Cr2O7: 12
H2SO4: 3
C2H5OH: 8
Cr2(SO4)3: 24
K2SO4: 4
CH3COOH: 4
H2O: 8