A number of antacid tablets contain water-insoluble metal hydroxides such as milk of magnesia, Mg(OH)2. Stomach acid is about 0.10 mol/L HCl(aq). Calculate the number of millimetres of stomach acid that can be neutralized by 500 mg of Mg(OH)2 (s).
To calculate the number of milliliters of stomach acid that can be neutralized by 500 mg of Mg(OH)2, you need to determine the number of moles of Mg(OH)2 and HCl and then use their stoichiometric ratio.
1. Determine the molar mass of Mg(OH)2:
Mg = 24.31 g/mol
O = 16.00 g/mol (2 atoms)
H = 1.01 g/mol (2 atoms)
Molar mass of Mg(OH)2 = 24.31 + 2(16.00) + 2(1.01) = 58.33 g/mol
2. Convert the mass of Mg(OH)2 to moles:
Moles of Mg(OH)2 = mass (g) / molar mass (g/mol)
Moles of Mg(OH)2 = 500 mg / 1000 mg/g / 58.33 g/mol = 0.0086 mol
3. Use the balanced chemical equation to determine the stoichiometric ratio between Mg(OH)2 and HCl. From the equation below, 2 moles of HCl react with 1 mole of Mg(OH)2:
Mg(OH)2 (s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
4. Calculate the number of moles of HCl:
Moles of HCl = 0.0086 mol × (2 moles HCl / 1 mole Mg(OH)2)
Moles of HCl = 0.0172 mol
5. Calculate the volume of stomach acid based on the concentration of HCl:
Volume (L) = Moles of HCl / Concentration (mol/L)
Volume (L) = 0.0172 mol / 0.10 mol/L = 0.172 L
6. Convert the volume from liters to milliliters:
Volume (mL) = Volume (L) × 1000 mL/L
Volume (mL) = 0.172 L × 1000 mL/L = 172 mL
Therefore, 500 mg of Mg(OH)2 can neutralize approximately 172 milliliters of stomach acid.
To calculate the number of millimeters of stomach acid that can be neutralized by 500 mg of Mg(OH)2, we need to use stoichiometry and the balanced chemical equation for the neutralization reaction.
The balanced chemical equation for the neutralization reaction between Mg(OH)2 and HCl is as follows:
Mg(OH)2 (s) + 2HCl (aq) -> MgCl2 (aq) + 2H2O (l)
From the balanced chemical equation, we can see that one mole of Mg(OH)2 reacts with two moles of HCl. Therefore, the molar ratio between Mg(OH)2 and HCl is 1:2.
First, we need to calculate the number of moles of Mg(OH)2 present in 500 mg. The molar mass of Mg(OH)2 is:
Mg(OH)2: Mg (24.31 g/mol) + 2(O) (16.00 g/mol) + 2(H) (1.01 g/mol) = 58.33 g/mol
Number of moles of Mg(OH)2 = Mass / Molar mass
Number of moles of Mg(OH)2 = 500 mg / 58.33 g/mol = 0.00856 mol
Next, we can use the mole ratio to determine the number of moles of HCl that can be neutralized. Since the mole ratio is 1:2, we multiply the number of moles of Mg(OH)2 by 2 to get the number of moles of HCl:
Number of moles of HCl = 0.00856 mol * 2 = 0.0171 mol
Finally, to calculate the volume of stomach acid that can be neutralized, we need to multiply the number of moles of HCl by the molar volume of a gas at standard conditions. The molar volume of a gas at standard conditions is 22.4 L/mol.
Volume of HCl(aq) = Number of moles of HCl * Molar volume of a gas at standard conditions
Volume of HCl(aq) = 0.0171 mol * 22.4 L/mol = 0.383 L
Since the volume of stomach acid is given in millimeters, we need to convert liters to millimeters. There are 1000 milliliters (mL) in one liter.
Volume of HCl(aq) = 0.383 L * 1000 mL/L = 383 mL
Therefore, 500 mg of Mg(OH)2 can neutralize approximately 383 millimeters of stomach acid.