Answer to Question #118697 in Chemistry for aljoc

Question

A teaspoon of milk of magnesia contains 12.0 mg of magnesium hydroxide. What volume of 0.010 mol/L HCl in a person's stomach would be neutralized by this teaspoon of antacid?

Answer 1

The reaction of neutralization of magnesium hydroxide by hydrochloric acid is:

Mg(OH)₂ + 2HCl$\rightarrow$ MgCl₂ + 2H₂O.

Therefore, 1 mol of magnesium hydroxide reacts with 2 mol of hydrogen chloride:

$n(Mg(OH)_2) = \frac{n(HCl)}{2}$ .

The number of the moles of magnesium hydroxide can be calculated as its mass divided by its molar mass (58.32 g/mol):

$n(Mg(OH)_2) = \frac{m}{M} = \frac{12.0·10^{-3} \text{ g}}{58.32 \text{ g/mol}} = 2.058·10^{-4}\text{ mol}$.

The number of the moles of HCl is then the double of this value:

$n(HCl) = n(Mg(OH)_2) ·2 = 4.115·10^{-4}\text{ mol}$ .

Finally, the volume of the HCl solution would be neutralized is:

$V = \frac{n}{c} = \frac{4.115·10^{-4}\text{ mol}}{0.010\text{ mol/L}} = 41.2$ mL.

Answer: the volume of 0.010 mol/L HCl would be neutralized by a teaspoon of antacid is 41.2 mL.