CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : Three hundred gallons of a mixtu

CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : Three hundred gallons of a mixture containing 75.0 wt % ethanol and 25 wt % water (mixture specific gravity = 0.877) and a quantity of a 40.0 wt % ethanol - 60 wt % water mixture (specific gravity = 0.952) are blended to produce a mixture containing 60.0 wt % ethanol. The specific gravity of a substance is the ratio of density of a substance compared to the density of water. The symbol of weight percent is wt %. (a) Estimate the specific gravity of the 60 % mixture by assuming that y = mx c where y is wt % ethanol, x is mixture specific gravity. Values for m and c are constants. (b) Determine the required volume of the 40 % mixture.

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CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : Three hundred gallons of a mixture containing 75.0 w..

Answer / kangchuentat

CHEMICAL MATERIAL BALANCE – ANSWER 2.2 : (a) Let X = specific gravity of the 60 % mixture. By interpolation for a straight line with constant m as gradient, then (75 - 40) / (0.877 - 0.952) = (60 - 40) / (X - 0.952). Then 35 / (-0.075) = -466.67 = 20 / (X - 0.952). X = -0.0429 + 0.952 = 0.9091. (b) Let V = the required volume of the 40 % mixture. Mass of ethanol in 75 % mixture + Mass of ethanol in 40 % mixture = Mass of ethanol in 60 % mixture. Then, 300 (0.877) (0.75) + V (0.952) (0.4) = (V + 300) (0.9091) (0.6). Then 197.325 + 0.3808 V = 0.5455 V + 163.638. V = 204.53 gallons.

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