Fluoride Ions needed to treat water

[MellowOne]

Homework Statement

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F - per 70 kg body weight can cause death. Nevertheless, in order to prevent tooth decay, F - ions are added to drinking water at a concentration of 1 mg of F - ion per L of water. How many kilograms of sodium fluoride is needed to treat a 7.50 x 10^2 gallon reservoir?

Homework Equations

1mg = 1.0 x 10^-6kg
1 gal = 283,905,884 liters

The Attempt at a Solution

I did 7.50 x 10^2 gallons and converted it to liters. Knowing how many liters is in the reservoir and knowing it takes 1.0 x 10^-6 kg to treat 1 liter of water, I got my answer as 284 kg. Anyone know what I did wrong? I'm fairly confident the answer is 284 kg.

 

[Borek]

1 gal = 283,905,884 liters

Doesn't look OK.

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[MellowOne]

I figured it out, I forgot a step and yeah I just looked at it again and realized I put the wrong conversion.