How many milliliters of 60% carbonic acid must be mixed with how many milliliters of 15% carbonic acid to make 650 milliliters of a 38% carbonic acid solution

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):