Please ck my math

Am I calculating the amount of Kmeta I am adding to rid the water of chlorine correctly?

I normally use distilled water and build from scratch therefore chlorine and chloramine are a non issue for me. However, the grocery has been unable to maintain a reliable stock of distilled water, I don’t operate a distillery, and I feel like those folks who require distilled water medically should get the limited supply on the shelves. So, that leaves me with New Market AL tap water.

Based on a photo of a package (link below), Campdem tablets = 550 mg KMeta which will treat 20 gal of chlorinated water. 550/20=27.5 mg per gallon. 1000 mg = 1 gram. So, .0275 grams per gal x 8.5 gal = .23 grams. I round up to .3

Edit:  based on post #6 below, in retrospect I imagine rounding up was a mistake.

Seems right to me

Yeah, the math is fine. You are treating 8.5 gallons instead of 20, so you only need 8.5/20 of the 0.55 gram tablet. 0.55*(8.5/20) = 0.23. I would round it up to 0.275 and use a half tablet.

[emoji106] I have powdered Kmeta so I’m trying to get a weight. I may round up a bit though.

Cheers!

The powder is more active, as there is no binder. Look at the low oxygen page for more info.

I may have learned that the hard way. In another thread I am dangling a 1967 penny in my beer to rid the evil from it thru alchemy.