How much water should be mixed with 6 gallons of 55% solution of sulfuric acid to make a 20% solution of sulfuric acid

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asked Oct 18, 2019 184k views

1 Answer

PKul · Answer 1 · 2019-10-23T12:35:42+0000

Hello.

For all dilution problems use the dilution equation:
C1V1 = C2V2
20*V1 = 6*55
V1 = 6*55/20
V1 = 16.5 gallons
The final volume must be 16.5 gallons
You must add 16.5 - 6 = 10.5 gallons water to be added.

Have a nice day

How much water should be mixed with 6 gallons of 55% solution of sulfuric acid to make a 20% solution of sulfuric acid

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