A diluted sample of a spirit is added to 50.00 mL of 0.0200 mol/L potassium dichromate and acid…

Step 1. Added. $50.00 \times 0.0200 = 1.000\ \text{mmol}$.

Step 2. Left over. Iron(II) used: $21.40 \times 0.1000 = 2.140\ \text{mmol}$. One dichromate takes six: $\dfrac{2.140}{6} = 0.357\ \text{mmol}$.

Step 3. Reacted. $1.000 - 0.357 = 0.643\ \text{mmol}$ of dichromate oxidised the ethanol.

Step 4. Ethanol. Three ethanol for two dichromate: $0.643 \times \dfrac{3}{2} = 0.965\ \text{mmol}$. Mass: $0.965 \times 46.069 = 44.5\ \text{mg}$.