Question

The approximate concentration of hydrochloric acid, HCl, in the stomach (stomach acid) is 0.17M. Calculate the mass of the following antacids required to neutralize 50mL of this acid:

a. Bicarbonate of soda, NaHCO3
b. Aluminum hydroxide, Al(OH)3

Answer 1

To calculate the mass of antacids required to neutralize 50 mL of 0.17 M hydrochloric acid, you need to determine the amount of acid neutralized by each antacid. For bicarbonate of soda (NaHCO3), 0.0085 moles of HCl is neutralized. For aluminum hydroxide (Al(OH)3), 0.0255 moles of HCl is neutralized.

Step-by-step explanation:

To calculate the mass of antacids required to neutralize 50 mL of 0.17 M hydrochloric acid, you need to determine the amount of acid neutralized by each antacid.

Let's calculate it for both antacids: a) Bicarbonate of soda (NaHCO3):

The balanced equation for the reaction between HCl and NaHCO3 is: HCl + NaHCO3 → NaCl + H2O + CO2. From the equation, 1 mole of HCl reacts with 1 mole of NaHCO3.

Given that the concentration of HCl is 0.17 M and the volume is 50 mL, we can calculate the number of moles of HCl using the formula: Moles = Concentration x Volume.

b) Aluminum hydroxide (Al(OH)3): The balanced equation for the reaction between HCl and Al(OH)3 is: 3HCl + Al(OH)3 → 3H2O + AlCl3. From the equation, 3 moles of HCl react with 1 mole of Al(OH)3.

Given that the concentration of HCl is 0.17 M and the volume is 50 mL, we can calculate the number of moles of HCl using the formula: Moles = Concentration x Volume.