SOLUTION: I am needing help figuring out this problem. I have tried several ways but can not figure it out. "A liquid that is 55% muriatic acid is added to 4 L of liquid that is 80% muriati

Question 786889: I am needing help figuring out this problem. I have tried several ways but can not figure it out.
"A liquid that is 55% muriatic acid is added to 4 L of liquid that is 80% muriatic acid how many liters of the 55% solution must be used to produce a new liquid that is 65% muriatic acid"
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
As a chemist, I would advice you not to do that. The solution will get hot, may boil over violently, and you could get burnt. Luckily this is just a math problem, and science does not apply, so we can assume that it can be done and that volumes are additive, too.

= amount of 55% muriatic acid added, in liters.
The amount of muriatic acid in that is (I won't even ask if it's in liters or kilograms).
The amount of muriatic acid in the 4 liters of 80% muriatic acid is
(in the same L or kg units).
The total amount of muriatic acid in the final mixture will be
.
The final volume (in math class, if not in real life) will be
liters.
Since those liters of final solution will be 65% muriatic acid, we can also calculate the amount of muriatic acid in the final mixture as

So,

Solving :

and multiplying both sides by 10 (or dividing both by 0.1, same thing)

So you are expected to add the 4L of 80% acid to of 55% acid.

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