SOLUTION: mixed 2 liters of 80% acid solution with 6 liters of 20% acid solution. What was the percent of acid in resulting mixture?

Question 278492: mixed 2 liters of 80% acid solution with 6 liters of 20% acid solution. What was the percent of acid in resulting mixture?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=percent of acid in resulting mixure
Now we know that the percent of acid in the 80% solution (0.80*2)+the percent of acid in the 20% soluton(0.20*6)=the amount of acid in the resulting mixture (x*8). So our equation to solve is:
0.80*2+0.20*6=8x
1.6+1.2=8x
2.8=8x
x=0.35 or 35%
Hope this helps--ptaylor
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