Question

A solution of 10% sulfuric acid and one of 25% sulfuric acid are to be used to make 500 ml of 20% sulfuric acid. How many ml of each solution needs to be used in the mixture? (round answers to the nearest ml)

Answer 1

Explanation:

x ml of 10 % sulfuric acid+(500-x) of 25% sulfuric acid=500 ml of 25% sulfuric acid

`(10x)/(100) +(25(500-x))/(100) =(500*20)/(100) \\10x+12500-25x=10000\\-15x=-2500\\3x=500\\x=(500)/(3) =166(2)/(3) \\500-x=500-(500)/(3) =(1000)/(3) =333(1)/(3) \\166(2)/(3) ~ml~of~10%~%~sulfuric~acid.\\333(1)/(3)~ml~of~25 %~%~sulfuric~acid\\`

Answer 2

Explanation:

If x is the volume of the 10% sulfuric acid, and y is the volume of 25% sulfuric acid, then:

x + y = 500

0.10 x + 0.25 y = 0.20 (500)

Solve the system of equations with substitution or elimination. Using substitution:

y = 500 − x

0.10 x + 0.25 (500 − x) = 0.20 (500)

0.10 x + 125 − 0.25 x = 100

25 = 0.15 x

x = 167 mL

y = 333 mL

You need 167 mL of 10% sulfuric acid and 333 mL of 25% sulfuric acid.