Question 387536: A chemist has only 50% and 20% sulfuric acid on his laboratory. How much of each should be mixed to obtain 12 liters of a 30% acid solution? Answer by gwendolyn(128) (Show Source):
You can put this solution on YOUR website! Let
f = amount of 50% solution used
t = amount of 20% solution used.
The problem states that we want a total of 12 liters of solution, so:
f + t = 12
We solve for f (we could solve for t, but we picked one at random):
f = 12 - t
We can also form an equation based on the anount of acid in each term.
The amount of acid contributed by f and the amount of acid contributed
by t added together need to add up to the amount of acid in the final solution.
To do this, we convert each percentage to the equivalent decimal factor
(50% = .5; 20% = .2; 30% = .3)
So, in equation form:
.5f + .2t = (.3)*12
.5f + .2t = 3.6
Let's multiply the entire equation by 10 just so we can use whole numbers:
5f + 2t = 36
Now we substitute the value of f from the first equation into the second equation:
5(12 - t) + 2t = 36
60 - 5t + 2t = 36
60 - 3t = 36
Subtract 60 from both sides to isolate the term with a variable:
60 - 3t - 60 = 36 - 60
-3t = -24
Divide both sides by -3 to solve for t:
t = 8
And since f = 12 - t, we can solve for f:
f = 12 - 8
f = 4
