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160 imperial gallons of a mixture contain 70 wt% ethanol and 30% water (density = 0.893 g/cm³). A quantity of a 40 wt% ethanol-water mixture (density = 0.931 g/cm³) is blended with this to produce a mixture containing 60 wt% ethanol.

Tasks:
1. Draw a schematic of the process, labeling all streams.
2. Calculate the volume of the 40% ethanol-water mixture required, in liters.

Enter your numerical answer only (omit units).

Answer :

Final answer:

To create a 60 wt% ethanol mix, 170.7 litres of a 40 wt% ethanol-water mixture should be added to 160 imperial gallons of a 70 wt% ethanol-water mixture.

Explanation:

In this question, you're trying to find out how much of a 40 wt% ethanol-water mixture you need to add to 160 imperial gallons of a 70 wt% ethanol-water mixture to result in a mixture with 60 wt% ethanol. To do this, you will need to use the concept of mass balance.

First, convert the volume of the 70% mixture to litres: 160 gallons = 727.4 litres. Then, calculate the mass of ethanol in the 70% mixture: 0.70 * 0.893 g/cm³ * 727.4 litres = 457.9 kg. Let's denote the volume of the 40% mixture as X litres. The total mass of ethanol in the final mixture comes from the ethanol in both mixtures, so it is 0.60 * (0.893 g/cm³ * 727.4 + 0.931 g/cm³ * X).

Setting these equal and solving for X yields X = 170.7 litres. You need 170.7 litres of the 40% ethanol-water mixture to make a final mix that is 60 wt% ethanol.

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