Salted water
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It is known that for a hydrochloric acid flow rate of 15 tons per hour, at 36.5%, how many tons of alkali are required to achieve complete neutralization? Detailed calculation formulas and procedures
These two solutions are neutralized in an equimolar ratio; since the total moles of acid are known, the total moles of base consumed are also equal. Just divide this total by the molar concentration of the alkaline solution.
What kind of operation is this – using valuable acids and bases to prepare inexpensive saline?
This question leaves one at a loss for words:shutup: How can such a thing happen?
When preparing saltwater for acid-base neutralization, leaving aside the issue of the cost of acids and bases, have you considered how to remove the heat generated?
Everyone is taking it too seriously; it’s probably just a middle school chemistry problem. Next time when working on such problems, one should be more careful – at least ask how many tons are needed per hour to achieve neutralization, right? Or change it to: how many tons of alkali are needed for 15 tons of acid. Also, perhaps the teacher’s idea is to have students point out that there’s a problem with this question, and there are extra points for that
Since the concentration of the lye is not specified, it is hydrochloric acid at 15 t/h and 36.5 wt%; therefore, a sodium hydroxide solution at 15 t/h and 40 wt% is required for neutralization. As for whether there is 40% liquid caustic, it doesn’t matter; anyway, it’s not possible to mix them directly like that