An old antacid commercial claimed that each tablet of their product could neutra
ID: 553528 • Letter: A
Question
An old antacid commercial claimed that each tablet of their product could neutralize 47 times its mass in stomach acid. The active ingredient in the antacid tablet, NaAl(OH)2CO3, reacts with the HCl in stomach acid according to the balanced reaction here. NaAI(OH)2CO, + 4 HCl NaCl + AlCl3 + 3 H2O +CO2 How many moles of HCI can a 1.67 g antacid tablet neutralize if the tablet contains 0.399 g of the active ingredient? Number moles HCI If stomach acid has a concentration of 0.14 M HCI, determine how many times the tablet's own mass of stomach acid can be neutralized by the 1.67 g antacid tablet. Assume that stomach acid has the density of water, 1.00 g/mL Number times the mass of the tabletExplanation / Answer
Moles of NaAl (OH)2CO3 = mass of it / ( molar mass of NaAl(OH)2CO3)
= ( 0.399g) / ( 144 g/mol) = 0.00277 mol
HCl moles needed = 4 x NaAl(OH)2CO3 ( as per balanced equation)
= 4 x 0.00277 = 0.0111 mol
Thus HCl moles needed = 0.0111 mol
B) Molarityof HCl = moles of HCl / volume of HCl
0.14 M = 0.0111 mol / ( volume of HCl)
volume of HCl = 0.0792 L =79.2 ml
mass of HCl = volume x density = 792 ml x 1g/ml = 79.2 g
Mass of HCl / mass of tablet = ( 79.2g / 1.67 g) = 47 times
Thus mass of HCl neutralised is 47 times that of mass of tablet