SOLUTION: If I had 20,000L of 28% HCL how much water would I need to dilute it to 15% HCL

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Question 942918: If I had 20,000L of 28% HCL how much water would I need to dilute it to 15% HCL
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
x = liters of 28% HCL/water = liters of (100-28)% water/HCL
y = liters of pure water
z = liters of 15% HCL/water = liters of (100-15)% water/HCL
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x = liters of 28% HCL/water = liters of 72% water/HCL
y = liters of pure water
z = liters of 15% HCL/water = liters of 85% water/HCL
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x = 20000
0.72x + 1.00*y = 0.85(x + y)
0.72*20000 + 1.00*y = 0.85(20000 + y)
0.72*20000 + y = 0.85*20000 + 0.85*y
y - 0.85*y = 0.85*20000 - 0.72*20000
(1 - 0.85)*y = 0.85*20000 - 0.72*20000
0.15y = 0.85*20000 - 0.72*20000
y = (0.85*20000 - 0.72*20000)/0.15
(0.85*20000 - 0.72*20000)/0.15
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copy and paste the above expression in to this calculator:
https://sooeet.com/math/scientific-calculator.php
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y = 17333.3333333
y = liters of pure water = 17333.3
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