SOLUTION: How many liters of pure (100%) acid must be added to 20 liters of 30% solution to bring the total solution up to 50% ?
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Question 903227: How many liters of pure (100%) acid must be added to 20 liters of 30% solution to bring the total solution up to 50% ?
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many liters of pure (100%) acid must be added to 20 liters of 30% solution to bring the total solution up to 50% ?
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Equation:
acid + acid = acid
0.30*20 + 1*x = 0.50(20+x)
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30*20 + 100x = 50*20 + 50x
50x = 20*20
x = (2/5)20
x = 8 liters (amt. of pure acid needed)
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Cheers,
Stan H.
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Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
1*x+0.3*20=0.5(20+x)
1*x+6=10+0.5x
1x-0.5x=10-6
0.5x=4
x=8 gallons of 100% is added to 20 gallons of 30% to get 28 gallons at 50%
check
1*8+0.3*20=0.5(20+8)
8+6=0.5(28)
14=14
ok
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