What mass of HCl, in grams, is neutralized by a dose of milk of magnesia containing 3.26 g of Mg(OH)2?
To determine the mass of HCl neutralized by 3.26 g of Mg(OH)2, we first need to calculate the moles of Mg(OH)2 present in the dose of milk of magnesia.
1. Calculate the molar mass of Mg(OH)2:
Mg: 24.305 g/mol
O: 16.00 g/mol (oxygen is present twice in the formula)
H: 1.008 g/mol (hydrogen is present twice in the formula)
Molar Mass of Mg(OH)2 = 24.305 + 2(16.00 + 1.008) = 58.319 g/mol
2. Calculate the moles of Mg(OH)2 present in 3.26 g:
moles = mass / molar mass
moles = 3.26 g / 58.319 g/mol ≈ 0.056 moles
3. Since Mg(OH)2 reacts with HCl in a 1:2 ratio, the moles of HCl that can be neutralized by 3.26 g of Mg(OH)2 is twice the moles of Mg(OH)2:
moles of HCl neutralized = 2 * moles of Mg(OH)2
moles of HCl neutralized = 2 * 0.056 = 0.112 moles
4. Calculate the mass of HCl neutralized by 0.112 moles:
mass = moles * molar mass
mass = 0.112 moles * 36.46 g/mol (molar mass of HCl) ≈ 4.08 g
Therefore, 3.26 g of Mg(OH)2 can neutralize approximately 4.08 grams of HCl.