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andrew-mc [135]
4 years ago
8

A chemist has one solu6on that is 40% sulfuric acid and one that is 10% sulfuric acid. How much of each should she use to make 2

0 liters of a solu6on that is 28% sulfuric acid?
Chemistry
1 answer:
kipiarov [429]4 years ago
4 0

Answer:

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.

Explanation:

For first solution of sulfuric acid :

C₁ = 40% , V₁ = ?

For second solution of sulfuric acid :

C₂ = 10% , V₂ = ?

For the resultant solution of sulfuric acid:

C₃ = 28% , V₃ = 20L

Also,

<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)

Using

<u>C₁V₁ + C₂V₂ = C₃V₃</u>

<u>40×V₁ + 10×V₂ = 28×20</u>

So,

40V₁ + 10V₂ = 560........................................(2)

Solving 1 and 2 as:

V₂ = 20 - V₁

Applying in 2

40V₁ + 10(20 - V₁)  = 560

40V₁ + 200 - 10V₁ = 560

30V₁ = 360

<u>V₁ = 12 L</u>

So,

<u>V₂ = 20 - V₁ = 8L</u>

<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>

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4. Given the balanced equation: 2Na + S → Na₂S
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