SOLUTION: I can't figure this out: Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?
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Question 347694: I can't figure this out: Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?
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Equation:
acid + acid = acid
x + 0.10(54-x) = 0.20*54
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Multiply thru by 100 and solve for "x":
100x + 10*54 - 10x = 20*54
90x = 10*54
x = 6 liters (amt. of pure acid needed in the mixture)
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Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let 
represent the amount of 10% solution that you have. Then 
represents the amount of pure acid needed.
)
Solve for then subtract the value of from 54.
John
My calculator said it, I believe it, that settles it
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