SOLUTION: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?

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Question 278419: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?
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Equation:
acid + acid = acid
0.80x + 0.30(400-x) = 0.62*400
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Multiply thru by 100 to get:
80x + 30*400 - 30x = 62*400
50x = 32*400
x = 8*32
x = 256 L (amt. of 80% solution needed in the mixture)
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Cheers,
Stan H.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
.8*x+.3*(400-x)=.62*400
x=256 400-x=144
256 L of 80%
144 L of 30%