It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and

88 wt.% water. Two solutions are available, the first contains 5 wt.% ethanol, and the second contains 25 wt.% ethanol. How much of each solution are mixed to prepare the desired solution?

**Solution:****1. Ethanol balance****Input = output****A (5 /100) + B( 25/ 100 ) = M ( 12 /100 )****0.05 A + 0.25 B = 0.12 M****A = (150 − 0.25 B)/ 0.05 = 3000 − 5 𝐵 … … … . . (𝟏)**

**2. Water balance****Input = output****0.95 A + 0.75 B = 0.88 M = 0.88 1250 = 1100****0.95 A + 0.75 B = 1100 … … … . . (𝟐)**

**Substituting (1) in (2)****0.95(300-5 B) + 0.75 B = 1100****2850 – 4.75 + 0.75 B =1100****4 B =1750 …………………. B= 437.5 kg****Substituting B in (1) : A= 3000 – 5(437.5) = 812.5 kg**

**3. Checking: Total material balance, Input = A + B = 437.5 + 812.5 =1250 kg****Output = M= 1250 kg**