Answer to Question #325118 in General Chemistry for Horse1

Question

Stomach acid is approximately 0.137 M HCl. Using your average number of moles of CaCO3 per antacid tablet, what volume of stomach acid will a tablet neutralize?

Answer 1

2 mol HCl - 1 mol CaCO₃

x mol HCl - 4.955×10^-3 mol CaCO₃

x = 4.955×10^-3 × 2 = 0.00991 mol

V(sol) = n(HCl) / C(sol) = 0.00991 mol / 0.137 mol/l = 0.07233 l = 72.33 ml