Question

mol in titration M xV Gin L) (0.100 L NaOH used) in neutralizing the antacid: moles of HCI added to the antacid moles of HCl used in titration mol Mass of antacid weighed mol mol LAB 8 The Neutra lizing Power of Antacids ere ME: Wikesson 1. How many milliliters of 0.100 M NaOH is needed to neutralize 20.0 mg of CACO3? roquired to boil after dissolving the antacid tablet (CACO3)? of coe Produce

Answer 1

Q1. The balanced equation is :

CaCO₃ + 2 HCl $\rightarrow$ CaCl₂ + H₂O + CO₂

mass CaCO₃ = 20.0 mg = 20.0 x 10^-3 g

moles CaCO₃ = (mass CaCO₃) / (molar mass CaCO₃)

moles CaCO₃ = (20.0 x 10^-3 g) / (100.09 g/mol)

moles CaCO₃ = 2.00 x 10^-4 mol

moles HCl required = (moles CaCO₃) * (2 moles HCl / 1 mole CaCO₃)

moles HCl required = (2.00 x 10^-4 mol) * (2 / 1)

moles HCl required = (2.00 x 10^-4 mol) * 2

moles HCl required = 4.00 x 10^-4 mol

volume HCl required = (moles HCl required) / (concentration HCl)

volume HCl required = (4.00 x 10^-4 mol) / (0.100 M)

volume HCl required = 4.00 x 10^-3 L

volume HCl required = 4.00 mL