Alcohol by back titration

A sample of a drink is warmed with an excess of acidified dichromate, which oxidises the ethanol to ethanoic acid: $2\text{Cr}_2\text{O}_7^{2-} + 3\text{C}_2\text{H}_5\text{OH} + 16\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 3\text{CH}_3\text{COOH} + 11\text{H}_2\text{O}$. The dichromate that is left over is then titrated with iron(II), in the ratio 1 dichromate to 6 iron(II). You will find the ethanol in the sample from what was added and what was left.

The idea. The dichromate used on the ethanol is the dichromate added minus the dichromate left over. Three ethanol molecules react with two dichromate ions.

Your tasks.

1. Set added to the millimoles of dichromate added.
2. Set left to the millimoles of dichromate left over, from the iron(II) titre.
3. Set used to the millimoles of dichromate that reacted with the ethanol.
4. Set mg_ethanol to the milligrams of ethanol in the sample.
5. Set percent to the mass of ethanol as a percentage of the mass of the sample.