If a tablet contains 129 mg of Mg(OH)2 and 505 mg of CaCO3, how many moles of HCl would theoretically be neutralized?
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
mols Mg(OH)2 = 0.129/molar mass = ?
mols CaCl3 = 0.505/molar mass.
?mol Mg(OH)2 x 2 = mols HCl
? mol CaCO3 x 2 = mols HCl
Total mols HCl = sum of the above.
To find out the number of moles of HCl that would theoretically be neutralized, we need to first determine which reactant will react first based on the stoichiometry of the balanced equation. Then we can calculate the number of moles corresponding to that reactant.
To do this, we need to know the balanced equation between HCl and the antacids (Mg(OH)2 and CaCO3). Assuming the balanced equation is:
2HCl + Mg(OH)2 -> MgCl2 + 2H2O
1 mole of Mg(OH)2 reacts with 2 moles of HCl, which means the stoichiometric ratio is 1:2.
Let's calculate the number of moles of Mg(OH)2 in the tablet:
Molar mass of Mg(OH)2 = 24.31 g/mol (molar mass of Mg) + 2 * 16.00 g/mol (molar mass of O) + 2 * 1.01 g/mol (molar mass of H)
= 58.33 g/mol
Number of moles of Mg(OH)2 = Mass of Mg(OH)2 / Molar mass of Mg(OH)2
= 129 mg / 58.33 g/mol
= 0.00221 moles of Mg(OH)2
Using the stoichiometric ratio, we can determine the moles of HCl necessary to react with 0.00221 moles of Mg(OH)2:
Number of moles of HCl = 2 * Number of moles of Mg(OH)2
= 2 * 0.00221 moles
= 0.00442 moles of HCl
Therefore, approximately 0.00442 moles of HCl would theoretically be neutralized.