(Eide Problem 14.20) Dilute sulfuric acid (19% acid and the rest water) is required for activating car batteries. A tank of weak acid (12.5% acid and the rest water) is available. If 450 lbs of 78% concentrate acid is added to the tank to get the required 19% acid, how much of the 19% acid is now available?

Available mass of 19% acid after mixing is 4534.62 lb

Explanation:

Ww are given:

Available conc. acid = 78%

Available weak acid = 12.5%

Used mass of conc. acid = 450 lbs

We will first mix x lb of 12.5% acid with 450lb of 78% acid to get y lb of 19% acid.

Let's make use of the mass balance equation:

mass out = mass in

Therefore, we have:

x + 450 lb = y

making x subject of formula

= x= y - 450lb

Using the mass balance equation for the mass of acid, we have:

mass of acid out = mass of acid in

Therefore we have:

(12.5% of x)+(78%of 450 lb)=(19%ofy)

= 0.125x + 351 lb = 0.19y

Solving for x:

Substituting y-450 lb for x in the equation above, we have:

= 0.125(y-450lb)=0.19y - 351 lb

=0.125y- 56.25lb = 0.19y -351 lb

= 0.125y - 0.19y = -351 lb + 56.25lb

y =4534.62 lb

catya that's not even a b or c