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

1 answer:

6 0

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

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