Question

A chemist has three different acid solutions. The first acid solution contains 20 % acid, the second contains 40 % and the third contains 70 % . They want to use all three solutions to obtain a mixture of 56 liters containing 50 % acid, using 2 times as much of the 70 % solution as the 40 % solution. How many liters of each solution should be used?

Answer 1

Answer:

The volume of the first acid solution (containing 20% acid) in the mixture is 14 liters

The volume of the second acid solution (containing 40% acid) in the mixture is 14 liters

The volume of the third acid solution (containing 70% acid) in the mixture is 28 liters

Step-by-step explanation:

The percentage of acid in the first acid solution = 20%

The percentage of acid in the second acid solution = 40%

The percentage of acid in the third acid solution = 70%

The volume they want to use the three solutions to obtain a mixture of = 56 liters

The concentration of the mixture they want to obtain = 50%

The amount of the 70% solution to be used = 2 × The volume of the 40% solution to be included

Let 'x', 'y', and 'z', represent the volumes of the first second and third acid solutions in the mixture

Therefore, from the question parameters, we get;

x + y + z = 56...(1)

z = 2·y...(2)

0.2·x + 0.4·y + 0.7·z = 0.5 × 56...(3)

Multiplying equation (3) by 10 and then subtracting equation (1) multiplied by 2 from the result gives;

10 × (0.2·x + 0.4·y + 0.7·z = 0.5 × 56) = 2·x + 4·y + 7·z = 5 × 56 = 280

2·x - 2·x + 4·y - 2·y + 7·z - 2·y = 280 - 2 × 56 = 168

2·y + 5·z = 168

From equation (2) and (4), we get;

2·y + 5·z = 2·y + 5×(2·y) = 168

12·y = 168

y = 168/12 = 14

The volume of the second acid solution in the mixture, y = 14 liters

From equation (2), we have;

z = 2·y

∴ z = 2 × 14 = 28

The volume of the third acid solution in the mixture, z = 28 liters

From equation (1), we get;

x + y + z = 56

x = 56 - (y + z)

∴ x = 56 - (14 + 28) = 14

The volume of the first acid solution in the mixture, x = 14 liters.