SOLUTION: pure acid is to be added to a 10% acid solution to obtain 42 L of a 40% acid solution. What amounts of each should be used? blank L of pure acid solution and blank L of 10% acid

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Question 534337: pure acid is to be added to a 10% acid solution to obtain 42 L of a 40% acid solution. What amounts of each should be used?
blank L of pure acid solution and blank L of 10% acid solution
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
x = amount 10% acid
y = amount of pure acid
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x+y = 42 liters
x = 42-y
y = 42-x
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The desired solution is 42 liters of a 40% solution. It will contain 42*.40 liters of pure acid.
42*.4 = 16.8 liters of pure acid and 25.2 liters of solvent (probably water).
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10%*x + 100%*y = 40%*42
.1*x + y = 16.8
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multiply both sides by 10 to eliminate the decimals
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x + 10y = 168
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substitute for x = 42-y
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42-y +10y = 168
9y = 126
y = 126/9 = 42/3 = 14
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x = 42-14
x = 28
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Check how much 'pure' acid is in the solution.
28 * 10% acid = 2.8 liters of pure acid
13 * 100% acid = 14 liters of pure acid
2.8 + 14 = 16.8 liters of pure acid in the 42 liters of solution
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Answer: Mix 14 liters of pure acid and 28 liters 10% acid to make 42 liters of 40% acid.
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Done.