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stomach acid is primarily an aqueous solution of .12 M hydrochloride acid. How many mL of stomach acid will be neutralized by two 1.25?
stomach acid is primarily an aqueous solution of .12 M hydrochloride acid. How many mL of stomach acid will be neutralized by two 1.25 gram tablet of alternately, assuming that the tablets are %95 aluminum hydroxide.
2 Answers
- hcbiochemLv 75 years agoFavorite Answer
Balanced equation: Al(OH)3 + 3 HCl --> AlCl3 + 3 H2O
Calculate the mass and moles of Al(OH)3
Mass Al(OH)3 = 2(1.25 g) X 0.95 = 2.38 g Al(OH)3
mol Al(OH)3 = 2.38 g / 78.00 g/mol = 0.0304 moles Al(OH)3
Moles HCl neutralized = 0.0304 mol Al(OH)3 X (3 mol HCl / 1 mol Al(OH)3) = 0.0913 mol HCl
Volume HCl neutralized = 0.0913 mol X (1 L/0.12 mol) = 0.761 L X 1000 mL/L = 761 mL stomach acid
- ?Lv 75 years ago
Al(OH)3 + 3HCl = AlCl3 + 3H2O
No. of Moles of Al(OH)3 = 1/3 x No. of moles of HCl
(2 x 1.25)g/78g/mole = 0.12M x V (where V = vol. of HCl.)
0.03205 mole = 0.12 x V
V = 0.03205/0.12 = 0.267L = 267mL. This is on assumption that the Al(OH)3 tablets were 100% pure.
If they are 95% pure the volume of acid needed is 95% of 267mL = 254ml.
