Question

The concentrations of tank 1 are 2 M hexane, 4M pentane and 2M isopropanol (M = moles/liter). Tank 2 has a composition of 1.5M decane, 3M isopropanol and 5.5M hexane. Tank 3 has a composition of 0.5M decane, 6M isopropanol and 6M hexane. You want to prepare 75 liters of a mixture that contains 3.5M hexane and 3M isopropanol. How much must you use of each tank?

Answer 1

Hexane from tank 1 is 2M and tank 2 is 5.5 M and tank 3 is 6M.

We use the formula  M final x V final = ( M1V1 + M2V2 + M3V3)

here M final is 3.5 M , V final = 75 L , M1 = 2M , M2 = 5.5 M and M3 = 6M

V1+V2+V3 - 75 L , hence V3 = 75 - V1-V2

substituting we get 3.5 x 75 = ( 2 x V1 + 5.5V2 + ( 6) ( 75-V1-V2)  

262.5 = - 4V1 - 0.5 V2 + 450

4V1 + 0.5V2 = 187.5 ........................(1)

Isopropanol from tank 1 is 2M and tank 2 is 3M and tank 3 is 6M. Required final Molarity 3M

Using the same formula which we used before

3 x 75 = ( 2 x V1) + ( 3 x V2) + ( 6) ( 75-V1-V2)

225 = -4V1 -3V2 + 450

4V1 + 3V2 = 225 ......................(2)

solving (1) and (2) we get V2 = 15 L , V1 = 45 L , and hence V3 = 15 L

Thus we use 15 L from tank 1 and 45 L from tank2 and 15L from tank 3