Gastric juice contains 3 gms of HCL per litre. If a person produces 2.5 litre of gastric juice per day, how many antacid tablets each containing 400 mg of Al(OH)3 are needed to neutralise all the acid produced in one day
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2.5L of gastric juice will produce 3*2.5 = 7.5g of HCl.
no of moles of HCl in 7.5g of HCl = 7.5/36.5 =0.205. No of moles in HCl = no of moles of H+ = 0.205
Thus we need 0.205 moles of OH- to neutralise.
Al(OH)3 tablet weighs 400mg = 0.4g
No.of moles of Al(OH)3 = 0.4/78 = 0.005
1 mole of Al(OH)3 produces 3 moles of OH- thus no of moles of OH- is 0.015
Thus no.of tables required to neutralise acid = 0.205/0.015 = 13.6 = 17 tables approximately
no of moles of HCl in 7.5g of HCl = 7.5/36.5 =0.205. No of moles in HCl = no of moles of H+ = 0.205
Thus we need 0.205 moles of OH- to neutralise.
Al(OH)3 tablet weighs 400mg = 0.4g
No.of moles of Al(OH)3 = 0.4/78 = 0.005
1 mole of Al(OH)3 produces 3 moles of OH- thus no of moles of OH- is 0.015
Thus no.of tables required to neutralise acid = 0.205/0.015 = 13.6 = 17 tables approximately
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