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A chemist has two solutions of H2SO4. One has a 40% concentration and the other has a 25% concentration.

How many liters of each solution must be mixed to obtain 78 liters of a 28% solution?

liters of the 40% solution and

liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.

(Round to the nearest tenth, if necessary.)

User Yeggeps
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Answer:

16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Step-by-step explanation:

Let the required volume of solution 1 be represented by x.

The required volume of solution 2 would then be 78-x.

The number of moles of solution 1 that would be required = 0.4x

The number of moles of solution 2 that would be required = 0.25(78-x)

The number of moles of the final mixture = 78 x 0.28 = 21.84

moles of solution 1 + moles of solution 2 = moles of final mixture

0.4x + 0.25(78 - x) = 21.84

0.4x + 19.5 - 0.25x = 21.84

0.4x - 0.25x = 21.84 - 19.5

0.15x = 2.34

x = 15.6 liters

To the nearest tenth = 16 liters

Liters of 40% solution needed = 16 liters

Liters of 25% solution needed = 78 - 16 = 62 liters.

Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

User BinaryCat
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