Gastric juice contains 3.0g of HCl per litre.If a person produces 2.5l of gastric juice per day,how many antacid tablets each containing 400mg of Al(OH)3 are needed to neutralized all the HCl produced in one day ?
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Answered by
24
Thanks for the question!
It is definitely a very interesting question to solve and do some brainstorming.
**************************************************NHCl in gastric juice =336.5
Meq of HCl =336.5×2500 ⇒205.48
w78/3×1000=205.48
WAl(OH)3=5.342g=5342mg
No of tablets Al(OH)3=5342400 ⇒13.34 ⇒≈14
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Hope it helps and solves your query!!
Answered by
57
1L gastric juice → 3.0g HCl
2.5L gastric juice → (3×2.50) = 7.50g HCl = 7.5/36.5 moles
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
3 moles of HCl would require 1 mole of Al(OH)₃. 7.5/36.5 moles of HCl require 7.5/36.5×1/3 moles of Al(OH)₃.
7.5/36.5×1/3 moles of Al(OH)₃ would weight 7.5/36.5×1/3×78 = 5.34g
Hence, number of tablets required to neutralise the Al(OH)₃ would be 5.34/.4 = 13.35 ≈ 14 tablets.
2.5L gastric juice → (3×2.50) = 7.50g HCl = 7.5/36.5 moles
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
3 moles of HCl would require 1 mole of Al(OH)₃. 7.5/36.5 moles of HCl require 7.5/36.5×1/3 moles of Al(OH)₃.
7.5/36.5×1/3 moles of Al(OH)₃ would weight 7.5/36.5×1/3×78 = 5.34g
Hence, number of tablets required to neutralise the Al(OH)₃ would be 5.34/.4 = 13.35 ≈ 14 tablets.
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