Question

How many grams of CaCO3 are required to neutralize 900. mL of stomach acid, which is equivalent to 0.0100 M HCl?

Answer 1

CaCO3 can neutralize stomach acid.

If stomach acid is equivalent to 900ml of 0.01 M HCl

the mmoles of HCl = molarity*volume in ml = 0.01M*900ml = 9 mmoles

The neutralization reaction will be -

CaCO3 + 2HCl -> H2O + CO2 + CaCl2

So, each mole of CaCO3 can neutralize 2 moles of HCl,

so, to neutralize 9 mmoles of HCl we need 9/2 mmoles of CaCO3

9/2 mmoles = 9/2 * 10^-3 moles

Molar mass of CaCO3 = molar mass of Ca + molar mass of C + molar mass of O*3

= 40g/mole + 12g/mole + 16g/mole*3 = 100g/mole

So, weight of 9/2 mmoles of CaCO3 = molar mass*moles

= 100g/mole * 9/2 * 10^-3 moles = 0.45 g

So, we need 0.45 g of CaCO3 to neutralize the stomach acid.