Not 100% sure, but I think 5 gallons.
20 gallons of 10% acid meaning 2 gallons of acid + 18 gallon of solvent (H2O/OH).
To get it down to 8% acid, you need to dilute it to 25 gallons because you know you have 2 gallons of acid, and 2/25 - 8%. So to get 25 gallons as your final volume, you would add 5 more gallons of the 10% OH.
Note that it doesn't really matter if it's 10% OH or 100% OH... and I'm assuming that the OH has no effect to the concentration of the acid (if it's base, then it would have an effect).
How do you solve for this:how many gallons of a 10% alcohol solution be mixed with 20 gallons of a 10% acid solution to obtain an 8% acid solution.
4 answers
how many gallons of a 10% alcohol solution be mixed with 20 gallons of a 10% acid solution to obtain an 8% acid solution?
This is college algebra not chemistry
This is college algebra not chemistry
The answer is 40/3 I just need to know how to solve it.
Strictly speaking, more information is needed for a truly correct solution due to molecular intermix between alcohol and water. In practical terms Terry is correct and his numbers are right and we will ignore the VERRY small error resulting. For the structured ALGEBRAIC solution see below
We have 20-gal of 10% acid and are going to dilute it with x-gal of 10% alcohol (presumably 0% acid) to yield (20+x)-gal of 8% acid.
--> 20(0.10)+x(0.00)=(20+x)(0.08)
--> 20(0.10)+0=20(0.08)+x(0.08)
--> 20(0.10-0.08)=x(0.08)
--> 20(0.02)=x(0.08)
X50 X50
--> 20=4x
/4 /4
--> 5=x --> x=5
I don't know where Krystal got 40/3 from but it does not work for the problem AS STATED.
Perhaps the original problem called for 5% acid instead of 10% alcohol
We have 20-gal of 10% acid and are going to dilute it with x-gal of 5% acid to yield (20+x)-gal of 8% acid.
--> 20(0.10)+x(0.05)=(20+x)(0.08)
--> 20(0.10)+x(0.05)=20(0.08)+x(0.08)
--> 20(0.10-0.08)=x(0.08-.05)
--> 20(0.02)=x(0.03)
X100 X100
--> 40=3x
/3 /3
--> 40/3=x --> x=40/3
We have 20-gal of 10% acid and are going to dilute it with x-gal of 10% alcohol (presumably 0% acid) to yield (20+x)-gal of 8% acid.
--> 20(0.10)+x(0.00)=(20+x)(0.08)
--> 20(0.10)+0=20(0.08)+x(0.08)
--> 20(0.10-0.08)=x(0.08)
--> 20(0.02)=x(0.08)
X50 X50
--> 20=4x
/4 /4
--> 5=x --> x=5
I don't know where Krystal got 40/3 from but it does not work for the problem AS STATED.
Perhaps the original problem called for 5% acid instead of 10% alcohol
We have 20-gal of 10% acid and are going to dilute it with x-gal of 5% acid to yield (20+x)-gal of 8% acid.
--> 20(0.10)+x(0.05)=(20+x)(0.08)
--> 20(0.10)+x(0.05)=20(0.08)+x(0.08)
--> 20(0.10-0.08)=x(0.08-.05)
--> 20(0.02)=x(0.03)
X100 X100
--> 40=3x
/3 /3
--> 40/3=x --> x=40/3