Question 5478: how much of a 25% acid solution should be mixed with a 50% acid solution to obtain 500 gallons of a 34%acid solution?
thanks for the help!
Answer by Abbey(339)   (Show Source):
You can put this solution on YOUR website! let the 25% acid solution quantity (volume) = x
Let y= volume of 50% acid solution.
because the total volume can equal only 500 gallons, we can say that 500 gallons - the quantity of 25% solution = the quantity of 50%, right? So:
500-x=y
To figure the solution we can multiply the percentage by the volume to have a "standard" to work with:
Let 25*x=25% standard
Let 50*y = 50% standard
Let 34*(500)=the final solution
This means we can build an equation that says, mix the amount of the 25% with an amount of the 50% and get 500 gallons of 34%:
25x+50y=34(500)
and now we can substitute a form of x in for our y value (because we know that 500-x=y) and solve for x:
25x+50y=34(500)
25x+50(500-x)=34(500)
25x+25000-50x=17000
-25x+25000=17000
25000=17000+25x
8000=25x
320=x
So we need 320 gallons of the 25% solution and 180 gallons of the 50% solution.
Check the results:
320 gallons *25% acid = 80 gallons of acid
180 gallons *50% acid = 90 gallons of acid
80+90=170
170 gallons/500 gallons = 34%
|
|
|
