The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.08 M K2Cr2O7 for titration. Should the police prosecute the individual for drunken driving if the legal limit is 0.1% by mass ethanol in the blood? The reaction is :CH3CH2OH + K2Cr2O7 + H2SO4 → CH3COOH + Cr2(SO4)3 + K2SO4 + H2O

Sure, let's solve this step by step.

1. First, we need to understand the stoichiometry of the reaction. From the balanced chemical equation, we can see that 1 mole of ethanol (CH3CH2OH) reacts with 1 mole of potassium dichromate (K2Cr2O7).

2. Next, we calculate the number of moles of K2Cr2O7 used in the titration. The number of moles is given by the formula:

   Moles = Molarity * Volume(L)

   So, Moles of K2Cr2O7 = 0.08 M * 4.23 mL = 0.08 M * 0.00423 L = 0.0003384 moles

3. Since the stoichiometry of the reaction is 1:1, the number of moles of ethanol in the blood sample is also 0.0003384 moles.

4. Now, we convert the moles of ethanol to grams. The molar mass of ethanol is approximately 46.07 g/mol, so:

   Mass of ethanol = Moles * Molar mass = 0.0003384 moles * 46.07 g/mol = 0.0156 g

5. The mass of the blood sample is given as 10.0 g. So, the percentage of ethanol by mass is:

   % ethanol = (Mass of ethanol / Mass of blood) * 100% = (0.0156 g / 10.0 g) * 100% = 0.156%

6. Comparing this with the legal limit of 0.1%, we can see that the alcohol content in the driver's blood is above the legal limit.

Therefore, the police should prosecute the individual for drunken driving.