Question

Four hundred fifty gallons (plus or minus a half gallon) of a mixture of 75.0 wt% ethanol and 25.0 wt% water (mixture specific gravity=0.877) and a quantity of 40.0 wt% ethanol and water (sg=0.952) are blended to produce a mixture containing 60.0 wt% EtOH. What is the volume (in gallons) of the 40% mixture required? How many pounds mass does the final solution consist of?

Answer 1

1 Gallon =3.78 Liters

Volume of mixture= 450 gallons= 450*3.78=1701 L

Specifc gravity = 0.877, specific gravity= density of given substance/ density of water

density of given substance =0.877*100= 877 kg/m3=877 g/L

So mass of mixture= 1701*877g/L=1491777gm =1491.777 Kg

Mass of ethanol= 1491.777*75/100=1118.833Kg mass of water = 1491.777- 1118.833=372.9433 Kg

Let x= mass of 40wt% ethanol that is mixed with   75 wt% ethanol

ethanol =0.4x mass of water= 0.6x

Mass of total mixture = 1491.777+x

mass of ethanol= 0.4x+1118.833

Mass % of ethanol is 60%, therefore 0.6*(1491.777+x)= 0.4x+1118.833

895 +0.6x= 0.4x+1118.833

0.2x= 1118.833-895=223.833 , x =1119.165 kg

So mass of final solution =(1491.777+1119.165)=2610.942 kg

1lb=0.4535 kg

2610.942 kg =2610.94/0.4535 lb=5757.134 lb

Mass of 755 ethanol to be added = 1119.165 kg

specific gravity of 75% ethanol =0. 952

Density of mixture/ density of water =0.952, density of mixture=0.952*1000 kg/m3=952 kg/m3=952 g/L

Volume= mass/ density= 1119.165*1000 /952 L=1175.593L

3.78 L= 1 Us Gallon

1175.593L    1175.593/3.78=311 Gallons