The approximate concentration of hydrochloric acid, in the stomach is 0.17M .calculate the mass of the following antacids required to neutralize 50mL of this acid;
1.) bicarbonate of soda, NaHCO3
2.) aluminum hydroxide, Al(OH)3
Pls answer
To calculate the mass of antacids required to neutralize hydrochloric acid, we need to determine the balanced chemical equations for the neutralization reactions and use stoichiometry to calculate the required amounts. Here are the steps to calculate the mass of each antacid:
1.) Bicarbonate of soda (NaHCO3):
The balanced chemical equation for the reaction between hydrochloric acid and sodium bicarbonate is:
NaHCO3 + HCl -> NaCl + CO2 + H2O
According to the balanced equation, the stoichiometric ratio between NaHCO3 and HCl is 1:1.
Step 1: Convert the given volume of HCl (50mL) to moles.
Molarity (M) = moles/volume (L)
Moles = M x Volume (in liters)
Given:
Concentration of HCl = 0.17M
Volume of HCl = 50 mL = 0.05 L
Moles of HCl = 0.17M x 0.05L = 0.0085 moles HCl
Step 2: Use the stoichiometric ratio to find the moles of NaHCO3 required.
Moles of NaHCO3 = Moles of HCl = 0.0085 moles NaHCO3
Step 3: Convert the moles of NaHCO3 to mass using its molar mass.
The molar mass of NaHCO3 = (22.99) + (1.01) + (12.01) + (3 x 16.00) = 84.01 g/mol
Mass of NaHCO3 = Moles of NaHCO3 x Molar mass of NaHCO3
Mass of NaHCO3 = 0.0085 moles x 84.01 g/mol
Mass of NaHCO3 ≈ 0.714 g
Therefore, approximately 0.714 grams of bicarbonate of soda (NaHCO3) is required to neutralize 50mL of the hydrochloric acid solution.
2.) Aluminum hydroxide (Al(OH)3):
The balanced chemical equation for the reaction between hydrochloric acid and aluminum hydroxide is:
3HCl + Al(OH)3 -> AlCl3 + 3H2O
According to the balanced equation, the stoichiometric ratio between Al(OH)3 and HCl is 1:3.
Step 1: Convert the given volume of HCl (50mL) to moles (as calculated in the previous example).
Moles of HCl = 0.0085 moles HCl
Step 2: Use the stoichiometric ratio to find the moles of Al(OH)3 required.
Moles of Al(OH)3 = Moles of HCl x (1 mole Al(OH)3 / 3 moles HCl)
Moles of Al(OH)3 = 0.0085 moles x (1/3) ≈ 0.0028 moles Al(OH)3
Step 3: Convert the moles of Al(OH)3 to mass using its molar mass.
The molar mass of Al(OH)3 = (1 x 27.00) + (3 x 16.00) + (3 x 1.01) = 78.00 g/mol
Mass of Al(OH)3 = Moles of Al(OH)3 x Molar mass of Al(OH)3
Mass of Al(OH)3 = 0.0028 moles x 78.00 g/mol
Mass of Al(OH)3 ≈ 0.218 g
Therefore, approximately 0.218 grams of aluminum hydroxide (Al(OH)3) is required to neutralize 50mL of the hydrochloric acid solution.