Question 163214

Let x= amount of pure acid needed
Now we know that the amount of pure acid in the 20% solution (0.20*12)plus tha amount of pure acid added (x) has to equal the amount of pure acid in the final mixture ((12+x)*0.45).  So our equation to solve is:
0.20*12+x=0.45(12+x) get rid of parens (distributive law) and simplify
2.4+x=5.4+0.45x  subtract 0.45x and also 2.4 from each side 
2.4-2.4+x-0.45x=5.4-2.4+0.45x-0.45x  collect like terms

0.55x=3  divide each side by 0.55
x=5.5 liters

CK
.20*12+5.5=0.45*17.5
2.4+5.5=7.875
7.9~~~7.875

Maybe an easier way:
Amount of otherstuff in the 20% solution (0.80*12) plus the amount of otherstuff in the pure acid added(0) has to equal the amount of otherstuff in the final mixture (0.55(12+x)).  So our equation to solve is:

0.80*12=0.55(12+x) get rid of parens and simplify
9.6=6.6+0.55x  subtract 6.6 from each side

9.6-6.6=0.55x  collect like terms
0.55x=3  -----------------same as before

Hope this helps---ptaylor