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Gastric juice contains 2.98 g HCL per liter. If a person produces about 2.5 L of...

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user9051586 | eNoter

Posted March 21, 2013 at 3:47 PM via web

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Gastric juice contains 2.98 g HCL per liter. If a person produces about 2.5 L of gastric juice per day, how many antacid tablets, each containing 400 mg of Al(OH)3 are needed to neutralize all the HCL produced in one day?

 

Al(OH3+HCl-->AlCl3+H2O

Balanced the reaction is Al(OH)3+3HCl--->AlCl3+3H2O

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted March 21, 2013 at 4:22 PM (Answer #1)

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`Al(OH)_3+3HCl rarr AlCl_3+3H_2O`

 

So to neutralize 3 HCl moles we need 1 mole of `Al(OH)_3` .

 

Amount of HCl in 1L of gastric juice `= 2.98g`

Amount of HCl in 2.5L of gastric juice `= 2.98xx2.5 = 7.45g`

 

So per day a person  will produce 7.45g of HCl.


Molic weight of HCl `= 36.5g/(mol)`

 

Amount of HCl moles produced per day `= 7.45/36.5`

Amount of HCl moles produced per day `= 0.204`

 

`Al(OH)_3:HCl = 1:3`

 

Amount of `Al(OH)_3` moles needed `= 0.204/3 = 0.068`

 

Molic weight of `Al(OH)_3 = 78g/(mol)`

 

Weight of `Al(OH)_3` needed to neutralize HCl ` `

`= 0.068xx78`

`= 5.304g `

`= 5304mg`

 

Amount of antacid tablets needed

`= 5304/400 `

`= 13.26`

`~= 14`

 

So the person need to get 14 tablets per day to neutralize the HCl acid.

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