Question 100199

Let x=amount of 50% solution that is produced

Then x-5=amount of 90% solution that was mixed with the distilled water

(1)  Pure sulfuric acid in the 90% solution=0.90(x-5)

(2)  Pure sulfuric acid in the distilled water=0

(3)  Pure sulfuric acid in the final mixture=0.50x

Now we know that (1)+(2)=(3).  So our equation to solve is:

0.90(x-5)=0.50x  get rid of parens

0.90x-4.5=0.50x  subtract 0.90x from both sides

0.90x-0.90x-4.5=0.50x-0.90x  collect like terms

-4.5=-0.40x  divide both sides by -0.40

x=11.25 liters -------------amount of 50% solution produced

CK

0.90*(11.25-5)=0.50*(11.25)
0.90(6.25)=5.625
5.625=5.625


Hope this helps---ptaylor