Question

How much of a solution that is 45% sulfuric acid needs to be mixed with a solution that is 90% sulfuric acid to produce 120 L of a solution that is 60% sulfuric acid

Answer 1

80 liters of 45% and 40 liters of 90% sulfuric acid solution

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Let's denote:

• the volume of the 45% solution as 'x' (in Liters)
• the volume of the 90% solution as 'y' (in Liters)

We have two key equations based on the question:

1. The total volume of the mixed solution should equal 120 L. So:
x + y = 120

2. The total volume of sulfuric acid in the mixed solution should be 60% of the total solution (which equals 60% of 120 L). Therefore:

• 0.45 * x (from the 45% solution) + 0.90 * y (from the 90% solution) = 0.60 * 120
  

Now we can solve this system of equations.

Starting with the first equation, we can express y in terms of x:

• y = 120 - x
  

Substituting this into the second equation gives us:

• 0.45x + 0.90 * (120 - x) = 0.60 * 120

Solving this for x, we find x = 80.

Therefore, we need 80 L of the 45% solution.

Now we can find the volume of the 90% solution 'y' by substituting x = 80 into the first equation:

• y = 120 - 80 = 40
  

Therefore, we need 40 L of the 90% solution.

So the answer is:

• 80 liters of the solution that is 45% sulfuric acid, and
• 40 liters of the solution that is 90% sulfuric acid.