Answer:
348.9 mL of the 60% solution and 251.1 mL of the 15% solution.
Explanation:
First, we calculate how many mililiters of pure carbonic acid are there in 650 mL of a 38% solution:
Then we can express the sum of both initial solutions as:
for the volume of carbonic acid; and
For the volume of the solutions.
We now have a system of two equations and two unknowns (x is the volume of the 60% solution and y is the volume of the 15% solution).
We express x in terms of y in equation 2):
And replace x in equation 1):
Finally we calculate x using equation 2):
