Question 572476
how much pure acid must be added to 6 gallons of 4% acid solution to make 20% acid solution? 
.
Let x = amount (gallons) of pure acid added
then
"pure acid" + "acid in 4% solution" = 20% of total mixture
x + .04(6) = .20(6+x)
x + .24 = 1.2+.20x
.80x + .24 = 1.2
.80x = 0.96
x = 1.2 gallons